Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
This project has been funded with support from the European Commission.
This publication reflects the views only of the author, and the Commission cannot be held
responsible for any use which may be made of the information contained therein.
Math E-learning
Materials
Gymnazium Teplice, Teplice, Czech Republic
Lyseopuistonlukio, Rovaniemi, Finland
The ZST Technical School, Mikolow, Poland
Gymnazium A. Bernolaka, Namestovo, Slovakia
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Description of the Project
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The main aim of our project „Math E-learning Materials“ is to prepare e-learning
materials which could be used at schools both by teachers and students. Produced
teaching units and materials can be used as an enrichment in order to provide engaging
interactive lessons blended with standard teaching methods. We would like to encourage
teachers to use blended learning methods. The produced materials can be found on the
project homepage, so it would be easily accessible also to students who can use them for
some extra practising.
Although our schools have different curricula, we have found what we have in
common. So we had decided to cover various topics (Algebra, Functions, Trigonometry
and Probability). The produced notebooks contain some theory, examples and graphs.
Students had a chance to work together and to get to know each other during local
project meetings in Mikolow - Poland, Teplice - Czech Republic, Graz - Austria and
Rovaniemi - Finland. This cooperation continued via emails, Skype and videoconferences.
They did work together on tasks given by the teachers, they exchanged their solutions
and discussed the problems.
Algebra
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Print
Linear equations and applications
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1.
Find four consecutive even integers such that the sum of the first three exceeds
the fourth by 8.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = the first even integer,
then x, x+2, x+4 and x+6 represent four consecutive even integers starting with the even
integer x.
The phrase "the sum of the first three exceeds the fourth by 8" translates into an equation:
Sum of the first three = fourth + excess.
x + (x+2) + (x+ 4) == (x + 6) + 8
Æ MDRealOnly ;
Input @
Clear@xD;
Solve@8x + Hx + 2L + Hx + 4L Hx + 6L + 8<, 8x< D
88x → 4<<
Answer: The four consecutive integers are 4, 6, 8 and 10.
2.
If one side of a triangle is one-third the perimeter, the second side is 7 meters,
and the third side is one-fifth the perimeter, what is the perimeter of the triangle?
MDPlotB80<, 8x, 0, 7<, PlotRange → 8−0.1, 7<,
AspectRatio → 1, Epilog → :[email protected], Point@80, 0<D,
Input @
Point@87, 0<D, [email protected], 2<D, Line@880, 0<, 82.5, 2<<D,
p
[email protected], 2<, 87, 0<<D, TextB" ", 85, 1.4<F,
3
p
TextB" ", 81, 1.3<F, Text@"7", 83.5, 0.4<D> F;
5
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1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
y
7
6
5
4
3
2
p
ÅÅÅÅÅ
3
p
ÅÅÅÅÅ
5
1
7
x
1
2
3
4
5
6
7
Let p = the perimeter.
Perimetr of triangle is p = a + b + c.
p p
Å + Å + 7= p
3 5
Æ MDRealOnly ;
Input @
Clear@pD;
p
p
+ 7 p>, 8p< F
SolveB: +
3
5
88p → 15<<
Answer: The perimeter is 15 meters.
3.
The distance along a shipping route between San Francisco and Honolulu is
2100 nautical miles. If one ship leaves San Francisco at the same time another
leaves Honolulu, and if the former travels at 15 knots and the latter at 20 knots,
how long will it take the two ships to rendezvous? How far will they be from
Honolulu and San Francisco at that time?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let t = number of hours until both ships meet
distance = speed x time
d1 = distance ship 1 from Honolulu travels to meeting point d1 = 20 t
d2 = distance ship 2 from San Francisco travels to meeting point d2 = 15 t
2 100 = total distance from Honolulu to San Francisco
d1 + d2 = 2100
20 t + 15 t = 2100
Æ MDRealOnly ;
Input @
Clear@tD;
Solve@8 20 t + 15 t == 2100 <, 8t < D
88t → 60<<
Input @
t = 60;
Input @
20 t
1200
Input @
15 t
900
Answer: It takes 60 hours for the ships to meet.
Distance from Honolulu is 1200 nautical miles.
Distance from San Francisco is 900 nautical miles.
4.
An excursion boat takes 1.5 times as long to go 360 miles up a river than to
return. If the boat cruises at 15 miles per hour in still water, what is the rate of
the current?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = rate of current (in miles per hour)
15 - x = rate of boat upstream
15 + x = rate of boat downstream
time upstream = (1,5) * (time downstream)
distance
time =
rate
distance upstream
rate upstream
360
15 - x
= 1,5 *
= H1, 5L *
distance downstream
rate downstream
360
15 + x
Æ MDRealOnly ;
Input @
Clear@xD;
360
SolveB:
== H1.5L
15 − x
360
15 + x
>, 8x< F
88x → 3.<<
Answer: The rate of the current is 3 miles per hour.
5.
An advertising firm has an old computer that can prepare a whole mailing in 6
hours. With the help of a newer model the job is complete in 2 hours. How long
would it take the newer model to do the job alone?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = time (in hours) for the newer model to do the job alone
(part of job completed in a given lingth of time) = (rate) * (time)
1
rate of old model = job per hour
6
rate of newer model =
1
job per hour
x
(part of job completed by old model in 2 hours) + (part of job completed by new model in 2
hours) = 1 whole job
(rate of old model) * (time of old model) + (rate of new model) * (time of new model) = 1
whole job
1
1
Å *2+ Å *2=1
6
x
Æ MDRealOnly ;
Input @
Clear@xD;
1
1
SolveB: 2 +
2 == 1>, 8x < F
6
x
88x → 3<<
Answer: The new computer could do the job alone in 3 hours.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
How many liters of a mixture containing 80% alcohol should be added to 5 liters
of a 20% solution to yield a 30% solution?
6.
Let x = amount of 80% solution used
(amount of alcohol in first solution) + (amount of alcohol in second solution) = (amount of
alcohol in mixture)
0,8 x + 0,2 * 5 = 0,3 * (x + 5)
Æ MDRealOnly ;
Input @
Clear@xD;
[email protected] x + 5 H0.2L == 0.3 Hx + 5L<, 8x< D
88x → 1.<<
Answer: Add 1 liter of the 80% solution.
Systems of linear equations and
applications
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7.
An individual wants to use milk and orange juice to increase the amount of
calcium and vitamin A in her daily diet. An ounce of milk contains 38 milligrams
of calcium and 56 micrograms of vitamin A. An ounce of orange juice contains 5
milligrams of calcium and 60 micrograms of vitamin A. How many ounces of milk
and orange juice should she drink each day to provide exactly 550 milligrams of
calcium and 1 200 micrograms of vitamin A?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = number of ounces of milk
y = number of ounces of orange juice
(calcium in x oz of milk) + (calcium in y oz of orange juice) = (total calcium needed
(milligrams))
38 x + 5 y = 550
(vitamin A in x oz of milk) + (vitamin A in y oz of orange juice) = (total vitamin A needed
(micrograms))
56 x + 60 y = 1 200
Æ MDRealOnly ;
Input @
Clear@xD;
NSolve@838 x + 5 y == 550, 56 x + 60 y == 1200<, 8x, y< D
88x → 13.5, y → 7.4<<
Answer: Drinking 13.5 ounces of milk and 7.4 ounces of orange juice each day will
provide the required amounts of calcium and vitamin A.
8.
An airplane makes the 2400 mile trip from Washington, D.C., to San Francisco
in 7.5 hours and makes the return trip in 6 hours. Assuming that the plane
travels at a constant airspeed and that the wind blows at a constant rate from
west to east, find the plane's airspeed and the wind rate.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = airspeed of the plane
y = the rate at which the wind is blowing
The ground speed of the plane is determined by combining these two rates:
x - y = ground speed flying east to west (headwind)
x + y = ground speed flying west to east (tailwind)
The formula of distance: d = r * t H d = distance, r = rate and t = time)
2 400 = 7.5 * (x - y) from Washington to San Francisco
2 400 = 6 * (x + y) from San Francisco to Washington
Æ MDRealOnly ;
Input @
Clear@x, yD;
Solve@82400 == 7.5 ∗ Hx − yL, 2400 == 6 ∗ Hx + yL<, 8x, y< D
88x → 360., y → 40.<<
Answer: The plane's airspeed is 360 mph and the wind rate is 40 mph.
9.
A publisher is planning to produce a new text-book. The fixed-costs (reviewing,
editing, typesetting, and so on) are $320 000, and the variable costs (printing,
sales commissions, and so on) are $31.25 per book. The wholesale price (the
amount received by the publisher) will be $43.75 per book. How many books
must the publisher sell to break even; that is, so that costs will equal revenues?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = the number of books printed and sold,
then the cost and revenue equations for the publisher are
y = 320 000 + 31.25 x cost equation
y = 43.75 x
The publisher breaks even when costs equal revenues.
Using the second equation to substitute for y in the first equation
320 000 + 31.25 x ã 43.75 x
Æ MDRealOnly ;
Input @
Clear@xD;
Solve@8320 000 + 31.25 x 43.75 x<, 8x<D
88x → 25 600.<<
Answer: The publisher wil break even when 25 600 books are printed and sold.
10.
Suppose you have $12 000 to invest. If part is invested at 10% and the rest at
15%, how much should be invested at each rate to yield 12% on the total
amount invested?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = the first part of investment
y = the second part of investment
x + y = invested amount = 12 000
0.1 (the thirst part of investment) + 0.15 (the second part of investment) =
(invested amount)
0.1 x + 0.15 y = 0.12 (x + y)
x + y = 12 000
0.12
Æ MDRealOnly ;
Input @
Clear@x, yD;
[email protected] x + 0.15 y == 0.12 Hx + yL, x + y == 12 000<, 8x, y< D
88x → 7200., y → 4800.<<
Answer: At 10% rate, $7 200 should be invested.
At 15% rate, $4 800 should be invested.
11.
A chemist has two solutions of hydrochloric acid in stock: a 50% solution and an
80% solution. How much of each should be used to obtain 100 mililiters of a
68% solution?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = amount of 50% solution
y = amount of 80% solution
x + y = 100 milliliters
68% = the target solution
0.5 x + 0.8 y = 0.68 (x + y)
Æ MDRealOnly ;
Input @
Clear@x, yD;
[email protected] x + 0.8 y == 0.68 Hx + yL, x + y == 100<, 8x, y<D
88x → 40., y → 60.<<
Answer: The chemist should use 40 ml of 50% solution and 60 ml of 80% solution.
12.
A fruit grower can use two types of fertilizer in an orange grove, brand A and
brand B. Each bag of brand A contains 8 pounds of nitrogen and 4 pounds of
phosphoric acid. Each bag of brand B contains 7 pounds of nitrogen and 7
pounds of phosphoric acid. Tests indicate that the grove needs 720 pounds of
nitrogen and 500 pounds of phosphoric acid. How many bags of each brand
should be used to provide the required amounts of nitrogen and phosphoric acid?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = the number of bags of brand A
y = the number of bags of brand B
one bag of brand A = 8 pounds of nitrogen and 4 pounds of phosphoric acid
one bag of brand B = 7 pounds of nitrogen and 7 pounds of phosphoric acid
amount of nitrogen = 720 pounds
amount of phosphoric acid = 500 pounds
8 x + 7 y = 720
4 x + 7 y = 500
Æ MDRealOnly ;
Input @
Clear@x, yD;
NSolve@88 x + 7 y == 720, 4 x + 7 y == 500<, 8x, y<D
88x → 55., y → 40.<<
Answer: 55 bags of brand A fertilizer and 40 bags of brand B fertilizer should be used.
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1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Quadratic equations and
applications
Open ê Close
13.
The sum of a number and its reciprocal is
13
. Find all such numbers.
6
Let x = the number
1
Å = the reciprocal number
x
(the number) + (the reciprocal number) =
13
6
1
13
x+ Å =
x
6
Æ MDRealOnly ;
Input @
Clear@xD;
SolveB:x +
::x →
2
3
1
x
==
>, :x →
3
2
13
6
>, 8x < F
>>
2
3
Answer: The numbers are Å and Å .
3
2
14.
An excursion boat takes 1.6 hours longer to go 36 km up a river than to return. If
the rate of the current is 4 km per hour, what is the rate of the boat in still water?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = rate of boat in still water
x + 4 = rate downstream
x - 4 = rate upstream
distance = 36 km
distance
the formula of time is
rate
(time upstream) - (time downstream) = 1.6
36
36
= 1,6
x-4 x +4
Input @
Clear@xD;
36
36
SolveB
−
== 1.6, 8x< F
x−4 x+4
88x → −14.<, 8x → 14.<<
Note: Because time can't have negative values, the first solution is excluded.
Answer: The rate in still water is 14 km/h.
15.
A payroll can be completed in 4 hours by two computers working simultaneously.
How many hours are required for each computer to complete the payroll alone if
the older model requires 3 hours longer than the newer model? Compute
answers to two decimal places.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = time for new model to complete the payroll alone
x + 3 = time for old model to complete the payroll alone
4 = time for both computers to complete the payroll together
1
= rate for new model
x
1
= rate for old model
x+3
(part of job completed by new model in 4 hours) + (part of job completed by old model in 4
hours) = 1 whole job
1
1
Å *4+
*4 =1
x
x+3
Æ MDRealOnly ;
Input @
Clear@xD;
1
1
SolveB: 4 +
4 == 1>, 8x<F
x
x+3
::x →
1
2
J5 −
73 N>, :x →
1
2
J5 +
73 N>>
Note: Because time can't have negative values, the first solution is excluded.
Input @
x=
1
2
J5 +
73 N êê N
6.772
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Tuesday, May 4, 2010
Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
x + 3
9.772
Answer: The new model would complete the payroll in 6.77 hours working alone, and the
old model would complete the payroll in 9.77 hours working alone.
16.
Find two numbers such that their sum is 21 and their product is 104.
Let x = the first number
y = the second number
sum of numbers is x + y = 21
product of numbers is x * y = 104
Æ MDRealOnly ;
Input @
Clear@x, yD;
Solve@8x + y == 21, x y == 104<, 8x, y< D
88x → 8, y → 13<, 8x → 13, y → 8<<
Answer: The numbers are 8 and 13.
17.
Find all numbers with the property that when the number is added to itself the
sum is the same as when the number is multiplied by itself.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = the number
sum of numbers is x + x
product of numbers is x * x
x+x=x*x
Æ MDRealOnly ;
Input @
Clear@xD;
Solve@8x + x == x x<, 8x< D
88x → 0<, 8x → 2<<
Answer: The numbers are 0 and 2.
18.
Find two consecutive positive even integers whose product is 168.
Let x = the first integer
x + 2 = the second integer
product of integers is 168
x * (x + 2) = 168
Æ MDRealOnly ;
Input @
Clear@xD;
Solve@8x Hx + 2L == 168<, 8x<D
88x → −14<, 8x → 12<<
Note: Only positive even integers.
Input @
x = 12
12
Input @
x+2
14
Answer: The integers are 12 and 14.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
If the length and width of a 4 m by 2 m rectangle are each increased by the
same amount, the area of the new rectangle will be twice that of the original.
What are the dimensions of the new rectangle (to two decimal places)?
MDPlotA80<, 8x, 0, 5.12<, PlotRange → 80, 3.12<,
Epilog → 9Purple, Line@880, 3.12<, 85.12, 3.12<<D,
Text@"4 + x", 82.5, 0.3<D, [email protected], 0<, 85.12, 3.12<<D,
Line@880, 12<, 815, 12<<D, TextA"S = H4 ∗ 2L∗2 m2 ",
Input @
82.5, 1.5<E, Text@"2 + x", 80.5, 2<D=E;
y
3
2.5
2
2+x
S = H4 * 2L*2 m2
1.5
1
0.5
4+x
x
1
2
3
4
5
Let 4 + x = the length of the new rectangle
2 + x = the width of the new rectangle
2 * 4 area of the rectangle
(2 * 4) * 2 area of the new rectangle
(x + 4) * (x + 2) = (2 * 4) * 2
Æ MDRealOnly ;
Input @
Clear@xD;
Solve@8Hx + 4L Hx + 2L == 2 × 4 × 2<, 8x< D
::x → −3 −
17 >, :x → −3 +
17 >>
Note: Becasue the length and the width can't have negative values, the first solution is
excluded.
Input @
x = −3 +
17 êê N
1.12311
Input @
x+4
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20
Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
5.12311
Input @
x+2
3.12311
Answer: The dimensions are 5.12 m and 3.12 m.
20.
Find the base b and height h of a triangle with an area of 2 square m if its base
1
is 3 m longer than its height and the formula for area is S = Å b.h.
2
MDPlotA80<, 8x, 0, 4<, PlotRange → 8−0.1, 2<,
Epilog → 9Purple, Line@880, 1<, 84, 0<<D,
Text@"h", 80.2, 0.5<D, Text@"b + 3", 81.5, 0.1<D,
TextA"S= 2m2 ", 81, 0.4<E, [email protected],
Input @
Point@84, 0<D, Point@80, 0<D, Point@80, 1<D=E;
y
2
1.5
1
0.5
h
S= 2m2
b+3
1
x
2
3
4
Let height h
base b = h + 3
area S = 2
1
area S = Å b * h
2
1
Å * (h + 3) * h = 2
2
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Æ MDRealOnly ;
Input @
Clear@hD;
1
SolveB: Hh + 3L h == 2>, 8h < F
2
88h → −4<, 8h → 1<<
Note: The height can't have negative values.
Input @
h=1
1
Input @
h+3
4
Answer: The base is 4 m and the height is 1 m.
21.
Two planes travel at right angles to each other after leaving the same airport at
the same time. One hour later they are 260 km apart. If one travels 140 km per
hour faster than the other, what is the rate of each?
MDPlot@80<, 8x, 0, 250<, PlotRange → 8−2, 102<,
Epilog → 8Purple, Line@880, 100<, 8240, 0<<D,
Text@"260 km", 8130, 60<D, Text@"x", 820, 50<D,
Input @
Text@"x + 140", 8100, 10<D, [email protected],
Point@80, 0<D, Point@8240, 0<D, Point@80, 100<D<D;
y
100
80
60
260 km
x
40
20
x + 140
x
50
Input @
100
150
200
250
2
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = rate of the first plane
x + 140 = rate of the second plane
distance = 260 km
Pythagorean theorem x2 + Hx + 140L2 = 2602
Æ MDRealOnly ;
Input @
Clear@xD;
SolveA9 x2 + Hx + 140L2 == 2602 =, 8x < E
88x → −240<, 8x → 100<<
Note: The rate can't have negative values.
Input @
x = 100
100
Input @
x + 140
240
Answer: The rates of the planes are 100 km/h and 240 km/h.
22.
One pipe can fill a tank in 5 hours less than another. Together they can fill the
tank in 5 hours. How long would it take to each alone to fill the tank? Compute
the answer to two decimal places.
Let x = time for first pipe to fill the tank alone
x + 5 = time for second pipe to fill the tank alone
5 = time for both pipes to fill the tank together
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
1
1
Å * 5+
* 5= 1
x
x+5
Æ MDRealOnly ;
Input @
Clear@xD;
1
1
SolveB:
5+
5 == 1 >, 8 x < F
x
x+5
::x →
5
J1 −
2
5 N>, :x →
5
2
J1 +
5 N>>
Note: The time can't have negative values.
Input @
x =
5
2
J1 +
5 N êê N
8.09017
Input @
x+5
13.0902
Answer: The faster pipe could fill the tank in 8.09 hours.
The slower pipe could fill the tank in 13.09 hours.
Systems involving second-degree
equations
Open ê Close
23.
An engineer is to design a rectangular computer screen with a 19 inch diagonal
and a 175 square-inch area. Find the dimensions of the screen to the nearest
tenth of an inch.
MDPlot@80<, 8x, 0, 15<, PlotRange → 80, 12<,
Epilog → 8Purple, Line@880, 12<, 815, 0<<D,
Input @
Text@"19 inch", 88, 7<D, Line@8815, 0<, 815, 12<<D,
Line@880, 12<, 815, 12<<D, Text@"S = 175 square−inch",
85, 4<D, Text@"x", 87, 1<D, Text@"y", 81, 7 <D < D;
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y
12
10
8
y
19 inch
6
4
S = 175 square-inch
2
x
x
2
4
6
8
10
12
14
Let x = the width
y = the height
19 = diagonal
area = 175 square-inch area
the formula for the area of a rectangle x * y = 175
Pythagorean theorem x 2 + y2 = 192
Æ MDRealOnly ;
Input @
Clear@x, yD;
NSolveA9x2 + y2 == 192 , x y 175=, 8x, y< E
88x → 11.674, y → 14.9906<, 8x → −11.674, y → −14.9906<,
8x → 14.9906, y → 11.674<, 8x → −14.9906, y → −11.674<<
Note: The dimension can't have negative values.
Input @
x = 14.990604021151215`
14.9906
Input @
y = 11.673979230795688`
11.674
Answer: The dimensions are 11.7 in and 15.0 in.
24.
Find two numbers such that their sum is 3 and their product is 1.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Let x = the first number
y = the second number
sum of numbers is x + y = 3
product of numbers is x * y = 1
Æ MDRealOnly ;
Input @
Clear@x, yD;
Solve@8x + y 3, x y 1<, 8x, y< D
::x →
:x →
1
2
1
2
J3 −
5 N, y →
J3 +
5 N, y →
1
2
1
2
1
Answer: The numbers are Å J3 +
2
25.
J3 +
5 N>,
J3 −
5 N>>
1
5 N and Å J3 2
5 N.
Find two numbers such that their difference is 1 and their product is 1. (Let x be
the larger number and y the smaller number.)
Let x = the first number
y = the second number
difference of numbers is x - y = 1
product of numbers is x * y = 1
Æ MDRealOnly ;
Input @
Clear@x, yD;
Solve@8x − y == 1, x y == 1<, 8x, y< D
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Tuesday, May 4, 2010
::x →
:x →
1
2
1
2
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
J1 −
5 N, y →
J1 +
5 N, y →
1
2
1
2
1
Answer: The numbers are Å J1 2
26.
J−1 −
5 N>,
J−1 +
5 N>>
1
5 N and Å J-1 2
1
5 N or Å J1 +
2
1
5 N and Å J-1 +
2
5 N.
Find the leghts of the legs of a right triangle with an area of 30 square meters if
its hypotenuse is 13 metres long.
MDPlotA80<, 8x, 0, 12<, PlotRange → 8−0.1, 5.1<,
Epilog → 9Purple, Line@880, 5<, 812, 0<<D,
Text@"13 m", 86, 3<D, TextA"S = 30 m2 ", 83, 2<E,
Input @
Text@"x", 85, 0.4<D, Text@"y", 80.6, 2.5<D, [email protected],
Point@80, 0<D, Point@812, 0<D, Point@80, 5<D =E;
y
5
4
3
13 m
y
2
S = 30 m2
1
x
x
2
4
6
8
10
12
Let
x = the length of the first leg of a right triangle
y = the length of the second leg of a right triangle
area of a right triangle 30 square meters
hypotenuse of a right triangle 13 meters
1
the formula for the area of a right triangle Å x * y = 30
2
Pythagorean theorem x 2 + y2 = 132
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Tuesday, May 4, 2010
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Æ MDRealOnly ;
Input @
Clear@x, yD;
SolveB:x2 + y2 == 132 ,
1
2
x y == 30>, 8x, y<F
88x → −12, y → −5<, 8x → −5, y → −12<, 8x → 5, y → 12<, 8x → 12, y → 5<<
Note: The lenght of the legs of a right triangle can't have negative values.
Answer: The lenghts are 5 meters and 12 meters.
27.
Find the dimensions of a rectangle with an area of 32 square meters if its
perimeter is 36 meters long.
MDPlotA80<, 8x, 0, 16<, PlotRange → 80, 2<,
Epilog → 9Purple, Line@880, 2<, 816, 0<<D,
TextA"S = 32 m2 ", 85, 0.8<E, Line@8816, 0<, 816, 2<<D,
Input @
Line@880, 2<, 816, 2<<D, Text@"O = 36 m", 88, 1.8<D,
Text@"x", 81, 1<D, Text@"y", 87, 0.2<D = E;
y
2
O = 36 m
1.75
1.5
1.25
1
x
S = 32 m2
0.75
0.5
0.25
y
x
2.5
5
7.5
10
12.5
15
Let
x = the first dimension of a rectangle
y = the second dimension of a rectangle
area of a rectangle 32 square meters
perimeter of a rectangle 36 meters
the formula for the area of a rectangle x * y = 32
the formula for the perimeter of a rectangle 2 x * 2 y = 36
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Æ MDRealOnly ;
Input @
Clear@x, yD;
Solve@8x y == 32, 2 x + 2 y == 36<, 8x, y<D
88x → 2, y → 16<, 8x → 16, y → 2<<
Answer: The dimensions are 2 m and 16 m.
28.
An artist is designing a logo for a business in the shape of a circle with an
inscribed rectangle. The diameter of the circle is 6.5 meters, and the area of the
rectangle is 15 square meters. Find the dimensions of the rectangle.
Clear@x, yD;
ImplicitPlotB:Hx − 3L2 + Hy − 1.25L2 6.5
2
>, 8x, −2, 7<,
2
PlotRange → 8−2, 5<, Epilog → 9Purple, Line@886, 0<, 86, 2.5<<D,
Input @
Text@"r = 6,5 m", 83.5, 1.8<D, Line@880, 2.5<, 86, 2.5<<D,
Text@"S", 83, 1<D, Line@880, 0<, 86, 2.5<<D, [email protected],
Point@83, 1.25<D, TextA"S = 15 m2 ", 81, 1.8<E,
Text@"x", 83, 0.3<D, Text@"y", 80.2, 1.2<D, Point@80, 0<D,
Point@86, 0<D, Point@86, 2.5<D, Point@80, 2.5<D=F;
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Tuesday, May 4, 2010
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y
5
4
3
2
1
S = 15 m2
r = 6,5 m
y
S
x
x
1
2
3
4
5
6
-1
-2
Let
x = the first dimension of a rectangle
y = the second dimension of a rectangle
6,5 meters is diameter of the circle and also of the rectangle
15 square meters is area of the rectangle
area of a rectangle x * y = 15
Pythagorean theorem x 2 + y2 = 6, 52
Æ MDRealOnly ;
Input @
Clear@x, yD;
SolveA9x2 + y2 == 6.52 , x y == 15=, 8x, y< E
88x → −6., y → −2.5<, 8x → −2.5, y → −6.<,
8x → 2.5, y → 6.<, 8x → 6., y → 2.5<<
Note: The dimensions of the rectangle can't have negative values.
Answer: The dimensions are 2.5 meters and 6 meters.
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Tuesday, May 4, 2010
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Functions
Open ê Close
Print
Linear Functions theory
Open ê Close
Definition of Linear Function
A function f is a linear function if f(x) = ax + b , for real numbers a and b.
The graph of a linear function is a straight line.
The domain of the function is all possible "x" values that satisfy the equation. The range
is all possible "y" values that satisfy the equation.
For example:
Input @
Clear@f, xD;
f@x_D = 2 x + 1;
Plot@8f@xD<, 8x, −3, 3<, PlotLegend → 8"f@xD"<,
LegendPosition → 81.1, −0.4<, PlotRange → 8−3, 3<,
Input @
AspectRatio → 1, ImageSize → 600D
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Tuesday, May 4, 2010
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y
3
2
1
-3
-2
-1
1
2
3
x
-1
-2
-3
The domain and the range of this fuction are all real number. D=R, Y=R
Zeros (Roots)
The zero of a function is the X-value for which Y, the value of the function is zero.
Input @
Clear@f, xD;
f@x_D = 3 x − 4;
Note: We can use FindRoot to calculate the root.
Input @
FindRoot@f@xD, 8x, 0<D
8x → 1.33333<
Note: We can also solve the equation f[x]=0. Now we can see that the root is x=1.(3).
Input @
MDPlot@8f@xD<, 8x, −2, 2<,
Epilog → 8Red, [email protected], [email protected], 0<D<,
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AspectRatio → 1, MDPlotLegend → True,
PlotRange → 8−2, 2<, ImageSize → 600D
y
2
1
-2
-1
1
2
x
-1
-2
Function monotonicity
Linear function is increasing function if a is positive.
Example:
Input @
Clear@f, xD;
f@x_D = 2 x + 1;
Input @
MDPlot@8f@xD<, 8x, −10, 10<, MDPlotLegend → True,
AspectRatio → 1, PlotRange → 8−10, 10<, ImageSize → 600D
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Tuesday, May 4, 2010
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y
10
5
-10
-5
5
10
x
-5
-10
Linear Function is decreasing if a is negative.
Example:
Input @
Clear@f, xD;
f@x_D = −2 x + 1;
Input @
MDPlot@8f@xD<, 8x, −10, 10<, AspectRatio → 1,
PlotRange → 8−10, 10<, MDPlotLegend → True, ImageSize → 600D
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Tuesday, May 4, 2010
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y
10
5
-10
-5
5
10
x
-5
-10
Linear function is constant function if a=0.
Example:
Input @
Clear@f, xD;
f@x_D = 1;
Input @
MDPlot@8f@xD<, 8x, −10, 10<, AspectRatio → 1,
PlotRange → 8−10, 10<, MDPlotLegend → True, ImageSize → 600D
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
y
10
5
-10
-5
5
-5
-10
'a' and 'b' coefficients
Coefficient 'a' of linear function says how line should be sloped to x-axis.
If 'a' is negative then function is decreasing. If 'a' is positive then function is increasing.
Clear@f, g, xD;
Input @ f@x_D = −5 x + 3;
g@x_D = 5 x + 3;
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Tuesday, May 4, 2010
Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
MDPlot@8f@xD, g@xD<, 8x, −5, 5<,
AspectRatio → 1, MDPlotLegend → True, ImageSize → 600D
y
20
10
-4
-2
2
4
x
-10
-20
Coefficient 'b' of linear function says how line should be moved on y-axis.
Clear@f, g, h, xD;
f@x_D = −5 x + 3;
Input @
g@x_D = −5 x + 6;
h@x_D = −5 x − 8;
Input @
MDPlot@8f@xD, g@xD, h@xD<, 8x, −6, 6<, AspectRatio → 1,
PlotRange → 8−6, 6<, MDPlotLegend → True, ImageSize → 700D
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Tuesday, May 4, 2010
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y
6
4
2
-6
-4
-2
2
4
6
-2
-4
-6
Note: If only 'b' coefficient is changed then lines are parallel.
Linear Functions examples
Open ê Close
1.
The store-house collected 21 tones of potatoes. Every day it gives 150
kilograms of potatoes. Draw a graph of the function defining the connection
between the number of kilograms of potatoes remaining in storage and time.
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Tuesday, May 4, 2010
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Note: We have to define function.
Input @
Clear@f, xD;
f@x_D = 21 000 − 150 x;
Note: Now we can plot the function.
Input @
MDPlot@8f@xD<, 8x, 0, 100<, ImageSize → 500D
y
20 000
18 000
16 000
14 000
12 000
10 000
8000
20
2.
40
60
80
From the swimming pool containig 1000m3 of water, 10m3 floats out each
minute. Show the volume of remaining water as the function of time t. Plot a
graph of this function.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Note: We have to define and plot the function.
Input @
Clear@f, tD;
f@t_D = 1000 − 10 t;
Input @
MDPlot@8f@tD<, 8t, 0, 100<,
AxesLabel → 8"t", "f@tD"<, ImageSize → 500D
f@tD
1000
800
600
400
200
20
3.
40
60
80
Metal rod in ambient temperature 0°C is 15m long. B y increasing the
temperature by 1°C the rod is 0,15mm longer. Show t he length of the rod as the
temprature function. In which ambient temperature the rod is 0.6cm longer?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Note: We have to define the function.
Input @
Clear@f, tD;
f@t_D = 0.015 t + 1500;
Note: t is the temperature of the rod, f[t] is the length of rod.
"Switch Solve, MDRealOnly, Pure Solve";
Input @ Clear@xD;
MDRealOnly@[email protected] f@tD<, 8t<DD
88t → 40.<<
Note: We can use Solve to find out the solution or we can plot the function and read
the solution.
MDPlot@8f@tD<, 8t, 10, 100<,
Input @
Epilog → 8Red, [email protected], Point@840, 1500.6<D<,
AxesLabel → 8 "t", "f@tD" <, ImageSize → 500D
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Tuesday, May 4, 2010
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f@tD
1501.4
1501.2
1501.0
1500.8
1500.6
1500.4
40
60
80
Answer: Metal rod will be 0.6cm longer in temeprature 40°C .
4.
From medical observations appears that time t[h] of human (aged to 18) daily
18 - n
where n is age in years. In which
sleep time is shown in the function t=8+
2
age man sleeps less than 10 hours per day, on average?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Note: We have to solve this equation.
Clear@t, nD;
Input @
SolveB:10 8 +
18 − n
2
>, 8n<F
88n → 14<<
Answer: Man after the age of 14 years sleeps on average less than 10 hours per day.
5.
In certain city, price of bus ticket is function of the number of traveled bus stops.
Write the function formula assuming that it is linear function and plot it for bus
line in which between ending and starting bus stop there are 18 other.
Clear@f, xD;
f@x_D = k ∗ x;
Input @
Clear@fTable, xD; fTable@x_D := x;
start = 0;
stop = 19;
Input @
step = 1;
data = Chop@Table@8x, N@fTable@xDD<, 8x, start, stop, step<DD;
MDShowTable@data, 8"x value", fTable@xD<D;
x value
x
0
1
2
3
4
5
6
0
1.
2.
3.
4.
5.
6.
7
8
9
10
7.
8.
9.
10.
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Tuesday, May 4, 2010
Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
11
12
13
11.
12.
13.
14
15
16
14.
15.
16.
17
18
19
17.
18.
19.
MDPlotData@data, FrameLabel → 8"X Values", "Y Values"<,
PlotStyle → 8Red, [email protected]<, PlotJoined → FalseD
19 Data Points
Y Values
15
10
5
0
0
5
10
15
X Values
Answer: This function is y=kx, for example y=x.
6.
The cost of hiring a store-house A can be counted as function k[x]=250+15x and
cost of hiring store-house B can be counted as function h[x]=90+45x, where x is
number of weeks of hiring. Make plots of these functions and find out which one
is more attractive and for how many weeks.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Note: We have to define both functions.
Clear@k, x, hD;
Input @ k@x_D = 250 + 15 x;
h@x_D = 90 + 45 x;
Note: Now we can plot both functions.
Input @
MDPlot@8k@xD, h@xD<, 8x, 0, 10<, AxesLabel → 8"weeks", "price"<,
MDPlotLegend → True, ImageSize → 700, PlotRange → 80, 700<D
price
700
600
500
400
300
200
100
0
2
4
6
8
10
Note: We should find the point where functions are crossing.
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Clear@xD;
N@Solve@8k@xD h@xD<, 8x<DD
88x → 5.33333<<
Answer: First store-house is more attractive to hire for longer than 5,3 weeks.
7.
Spring's length y[cm] is a function of weighting with mass x[kg] declared by
function y=2x+20. Allocate values of mass x, which you can weight spring, so
the length would be between 30cm and 70cm.
Note: We have to find out weight when length of the line is 30 and 70 cm long.
Input @
Clear@x, yD;
Solve@830 2 x + 20<, 8x<D
88x → 5<<
Input @
Solve@870 2 y + 20<, 8y<D
88y → 25<<
Answer: Weight should be between 5 and 25 kg.
8.
It is known that from 90kg of dry linum it is possible to make 8kg linen canvas.
How many kg of linum is needed to make 6kg linen canvas? Write a function
declaring amount of gained linen canvas depending on the amount of linum
taken to production?
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Note: We have to create proportion.
90kg 8kg
xkg
6kg
Clear@xD;
Input @
NBSolveB:x 90 × 6
8
>, 8x<FF
88x → 67.5<<
Answer: To make 6kg linen canvas we have to use 67.5 linum.
Note: We have two points A=(67.5, 6) and B=(90, 8).
xA =67.5
yA =6
xB =90
yB =8
Input @
xA = 67.5;
Input @
yA = 6;
Input @
xB = 90;
Input @
yB = 8;
Input @
Clear@x, yD;
Solve@8 Hy −yA L ∗ HxB −xA L == HyB −yA L ∗ Hx −xA L<, 8 y < D
88y → 0.0444444 H135. + 2. H−67.5 + xLL<<
Input @
FullSimplify@y = 0.044444444444444446` H135.` + 2.` H−67.5` + xLLD
0. + 0.0888889 x
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Tuesday, May 4, 2010
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
Input @
Clear@f, xD;
f@x_D = 0.08888888888888889 x;
Input @
MDPlot@8f@xD<, 8x, 0, 100<,
AxesLabel → 8 "x", "f@xD" <, ImageSize → 500D
f@xD
8
6
4
2
20
40
60
80
Answer: This function is f[x]=0.0(8)x.
Quadratic functions - theory
Open ê Close
Definition of a Quadratic Function
Quadratic functions are any functions that may be written in the form
y = ax2 + bx + c where a, b, and c are real coefficients and a ∫ 0.
The expression ax2 + bx + c is called 2nd degree polynomial.
Graph of a Quadratic Function
The graph of a quadratic function is called a parabola.
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Examples:
Input @
Clear@f, xD;
f@x_D = x2 ;
Input @
MDPlot@8f@xD<, 8x, −3, 3<, MDPlotLegend → True,
AspectRatio → 1, PlotRange → 8−3, 3<, ImageSize → 500D
y
3
2
1
-3
-2
-1
1
2
3
x
-1
-2
-3
Input @
Clear@f, xD;
f@x_D = −x2 ;
Input @
MDPlot@8f@xD<, 8x, −3, 3<, MDPlotLegend → True,
AspectRatio → 1, PlotRange → 8−3, 3<, ImageSize → 500D
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Tuesday, May 4, 2010
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y
3
2
1
-3
-2
-1
1
2
3
x
-1
-2
-3
Maximum, Minimum, Vertex of a
Quadratic Function
The Vertex is the highest (maximum) or lowest (minimum) point of the graph.The vertex is
the highest point on a parabola if the parabola opens down (a<0), it is the lowest point if
the parabola opens up (a>0).
Clear@f, g, h, xD;
f@x_D = x2 ;
Input @
g@x_D = 2 x2 + 3 x − 1;
h@x_D = −0.5 x2 − x + 3;
Input @
MDPlot@8f@xD, g@xD, h@xD<, 8x, −5, 5<, AspectRatio → 1,
PlotRange → 8−5, 12<, MDPlotLegend → True, ImageSize → 800D
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Tuesday, May 4, 2010
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y
10
5
-4
-2
2
4
-5
The x-coordinate of the vertex is x=
-b
. The y-coordinate of the vertex is y=-
D
. In
2a
4a
Mathematica we can use FindMinimum or FindMaximum depending on what we are
looking for.
Note: D=b2 -4ac
Input @
FindMaximum@h@xD, 8x<D
83.5, 8x → −1.<<
Note: Coordinates of vertex of function h[x] are (-1,3.5).
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Tuesday, May 4, 2010
Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
FindMinimum@g@xD, 8x<D
8−2.125, 8x → −0.75<<
Note: Coordinates of vertex of function g[x] are (-0.75,-2.125).
Input @
FindMinimum@f@xD, 8x<D
80., 8x → 0.<<
Note: Coordinates of vertex of function f[x] are (0,0).
MDPlot@8f@xD, g@xD, h@xD<, 8x, −5, 5<,
PlotLegend → 8"f@xD", "g@xD", "h@x"<, PlotRange → 8−4, 4<,
Input @
Epilog → 8Red, [email protected], Point@8−1, 3.5<D,
Point@8−0.75, −2.125<D, Point@80, 0<D<,
MDPlotLegend → True, ImageSize → 800D
y
4
2
-4
-2
2
4
-2
-4
Zeros
Function f[x] = ax2 + bx + c (a∫0):
1. For D>0 has 2 different roots: x1 =
Site: www.deltasoft.at
-b 2a
M@th Desktop
D
and x2 =
-b +
D
2a
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2. For D=0 has 1 (double) root x=
-b
2a
3. For D<0 has no roots
Examples:
Find roots of function f[x]=-2 x2 -x+6.
We have to calculate D first.
Clear@a, b, c, ∆, x1 , x2 D;
a = −2;
Input @
b = −1;
c = 6;
Input @
SolveA9∆ b2 − 4 a c=, 8∆<E
88∆ → 49<<
Input @
∆ = 49;
Note: Now we can calculate roots.
Input @
SolveB:x1 ::x1 →
3
2
−b −
∆
2a
, x2 −b +
2a
∆
>, 8x1 , x2 <F
, x2 → −2>>
3
Answer: Roots of function f[x] are x1 = Å , x2 = -2.
2
Forms of a quadratic function
1. f[x] = ax2 + bx + c is called the standard form.
2. f[x] = a(x-x1 )(x-x2 ) is called the factored form, where x1 and x2 are the roots of the
quadratic equation.
3. f[x] = aHx - pL2 +q is called the vertex form, where p and q are the x and y coordinates of
the vertex.
Transforming vertex form to standard form:
Example: Write y=3Hx - 2L2 +4 in standard form.
We just have to expand this equation.
Input @
ExpandA3 Hx − 2L2 + 4E
16 − 12 x + 3 x2
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Answer: Standard form is y=3 x2 -12x+16.
Transforming standard form to vertex form:
Example: Write y=x2 -6x+4 in vertex form.
We already know a=1. We just need to calculate p and q.
-b
D
p=
and q=(D=b2 -4ac)
2a
4a
Clear@a, b, c, p, q, ∆D;
a = 1;
Input @
b = −6;
c = 4;
Input @
SolveB:p −
b
2a
>, 8p<F
88p → 3<<
Input @
p = 3;
Note: p = 3. Now we need to calculate D.
Input @
SolveA9∆ b2 − 4 a c=, 8∆<E
88∆ → 20<<
Input @
∆ = 20;
Note: D = 20 so now we can calculate q.
Input @
SolveB:q −
∆
4a
>, 8q<F
88q → −5<<
Note: Now we know all needed variables to complete vertex form.
f[x]=aHx - pL2 +q
f[x]=Hx - 3L2 -5
Transforming factored form to standard form:
To do this we just have to use a "FullSimplify" on the equation, for example:
1
Write f[x]=- Å (x-3)(x+3) in standard form:
3
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Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
FullSimplifyB−
1
3
3−
Hx − 3L Hx + 3LF
x2
3
1
Answer: Standard form of this function is f[x]=- Å x2 +3
3
Transforming standard form to factored form:
We have to calculate roots and put them into formula (we already know a).
Example:
Write f[x]=5x2 - 15 x in factored form.
Clear@a, b, c, ∆, x1 , x2 D;
a = 5;
Input @
b = −15;
c = 0;
Input @
SolveA9∆ b2 − 4 a c=, 8∆<E
88∆ → 225<<
Input @
∆ = 225;
Note: Now we can calculate roots.
Input @
SolveB:x1 −b −
2a
∆
, x2 −b +
2a
∆
>, 8x1 , x2 <F
88x1 → 0, x2 → 3<<
Note: 5(x-0)(x-3)=5x(x-3)
Answer: Factored form of this function is f[x]=5x(x-3).
Quadratic functions
examples
-
Open ê Close
9.
The number of admissions to hospital on an n-day of the epidemy can be
presented in a formula A(n) = 10 +30n-3n2 . Find out how many sick people were
admissioned to hospital during one day.
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Note: First we will find maximium arrived patients in one day - it's a vertex of A(n)
function (so also maximum of that function)
Input @
Clear@f, nD;
f@n_D = 10 + 30 n − 3 n2 ;
Input @
FindMaximum@f@nD, 8n<D
885., 8n → 5.<<
Answer: The maximum amount of patients in one day was 85
10.
On the 9-th March, 2010 a person celebrating their jubilee said:
"When I multiply my age of 17 years earlier and my age in 20 years time, I'll get
this year, 2010". Which birthday is he/she celebrating?
Note: We must find a pattern first. x is a current age of a birthday
Input @
FullSimplify@Hx − 17L ∗ Hx + 20L 2010D
x H3 + xL 2350
Note: The pattern is x(3+x) = 2350. You can write it in form: f[x]=x2 +3x-2350. As you
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can see it is a quadratic function.
"Switch Solve, MDRealOnly, Pure Solve";
Input @ Clear@xD;
MDRealOnlyASolveA9x2 + 3 x − 2350 0=, 8x<EE
88x → −50<, 8x → 47<<
Note: As you see, we have 2 answers, but the first one (-50) is incorrect (we can't have
a minus age)
Answer: The birthday have 47 years
11.
There was a shop and its number of clients can be expressed by a formula d(n)
= -2n2 +32n-8, where n is the number of days, nœN+ and
1§n§15. Which day was the number of clients the biggest?
Note: First thing to do is define a function
Input @
Clear@d, nD;
d@n_D = −2 n2 + 32 n − 8;
Note: Secondly we will find a maximum
Input @
FindMaximum@d@nD, 8n<D
8120., 8n → 8.<<
Answer: Most customers came in 8th day.
A bullet moving upwards falls down verticaly after reaching the maximum point
of its route.Quadratic function can be used to describe it. For example the
2
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function h(t) = -5t2 +50t+5 assigns to time t[s] the height h[m] where the bullet is.
a) Plot the graph of that function.
b) What is the maximum height which bullet reached?
c) How long the bullet was in air?
d) After what time the bullet was in the same height which it was launched?
Note: For plot the data graph we will use 'def f' and 'Plot' buttons
Input @
Clear@h, tD;
h@t_D = −5 t2 + 50 t + 5;
Input @
MDPlot@8h@tD<, 8t, 0, 11<,
PlotRange → 80, 140<, AxesLabel → 8 "t", "h@tD" <D
h@tD
140
120
100
80
60
40
20
0
2
4
6
8
10
t
Note: For find the maximum height we will use a 'FindMaximum' function built-in M@th
Desktop
Input @
FindMaximum@h@tD, 8t<D
8130., 8t → 5.<<
Answer: The maximum height ehich bullet reached is 130 meters
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Note: for check how long the bullet was in air we will use a 'Solve' button
"Switch Solve, MDRealOnly, Pure Solve";
Input @ Clear@xD;
N@MDRealOnly@Solve@8h@tD 0<, 8t<DDD
88t → −0.0990195<, 8t → 10.099<<
Note: We have two answers - of course the first one (-0.0990195) is incorrect.
Answer: The bullet was in air 10.099 seconds
Note: To check from which height the bullet was launched we must check what value
the function have for 0 second
Input @
h@0D
5
Note: The bullet was launched from 5 meters. We will use now again a 'Solve' button to
check after what time the bullet reach the same height
"Switch Solve, MDRealOnly, Pure Solve";
Input @ Clear@xD;
MDRealOnly@Solve@8h@tD 5<, 8t<DD
88t → 0<, 8t → 10<<
Note: First answer we already have - it's the time when the bullet was launched.
Answer: Bullet reached the same height after 10 seconds
An electric train sets from station A, and stops at station B. It speed vB
13.
km
h
F, after
1
t minutes from leaving the station A is described by the formula: v= Å t H4 - tL.
5
Find out the time to cover the distance from station A to station B.
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Note: We will define a pattern
Clear@v, t, xD;
Input @
1
v@t_D =
t H4 − tL;
5
Note: To calculate the time of travel we will use a 'Solve' button.
Input @
"Switch Solve, MDRealOnly, Pure Solve";
MDRealOnly@Solve@8v@tD 0<, 8t<DD
88t → 0<, 8t → 4<<
Answer: The travel time from station A to B equals 4 minutes
14.
You are the cinema owner and you know that when the ticket price is 10 €, the
number of people coming to the cinema is about 100. When the price is
increased by one Euro, the number of clients is less by 5 people. Which price is
the best for your income?
Note: First we must find a pattern. The price of ticket is 10+x and the number of people
is 100 - 5x.
Input @
Clear@f, xD;
f@x_D = Expand@H10 + xL H100 − 5 xLD
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1000 + 50 x − 5 x2
Note: For a better look we can plot that function
Input @
MDPlot@8f@xD<, 8x, 0, 10<, AxesLabel → 8 "x", "f@xD" <D
f@xD
1120
1100
1080
1060
1040
1020
2
4
6
8
10
Note: Now we must find a maximum value of that function
Input @
FindMaximum@f@xD, 8x<D
81125., 8x → 5.<<
Answer: For the best income you should set the price of ticket to 5 €.
15.
A function defined by a formula f(x) =
-x2 + 6 x + 21
describes a worker's
2
productivity depending on working time x during eight hour working day. This
worker starts working at 7 a.m. When is the highest productivity?
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Note: We will define a pattern
Input @
Clear@f, xD;
1
f@x_D =
I−x2 + 6 x + 21M
2
1
2
Input @
1
2
I21 + 6 x − x2 M
I21 + 6 x − x2 M
Note: Now we will find a maximum
Input @
FindMaximum@f@xD, 8x<D
815., 8x → 3.<<
Note: The third hour of work is the most performance hour. So if worker starts his job at
7 a.m. then he reach the best performance in 10 a.m..
Answer: The performance of worker is the biggest at 10 a.m.
16.
66 games were played in a chess tournament, in which each player played one
game with each competitor. How many people took part in this competition?
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Note: First we must find a pattern. Let the x describe a number of all players.
Input @
FullSimplify@x Hx − 1L ê 2 66D
x2 132 + x
Note: As you can see it's a quadratic function: f(x) = x2 -x-132. Now we must find a
number of players, so we must
Clear@x1 , x2 D;
a = 1;
Input @
b = −1;
c = −132;
Input @
∆ = b2 − 4 ∗ a ∗ c
529
Input @
SolveB:x1 −b −
∆
2a
, x2 −b +
2a
∆
>, 8x1 , x2 <F
88x1 → −11, x2 → 12<<
Note: We have two answers but -11 is incorrect (there can't be -11 number of players
on a tournament)
17.
A life guard has got a rope and wants to make a special rectangular zone for
children with this rope of the biggest possible area. Give the dimensions for a
rope 100 meter long.
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Note: First we must find a pattern. Let the a and b will the sides of the rectangle
Input @
Clear@a, b, c, f, xD
Input @
2 a + 2 b 100;
Note: Then b=50-a
Input @
f@a_D = a H50 − aL
H50 − aL a
Note: Now we must find a maximum of that function
Input @
FindMaximum@f@cD, 8c<D
8625., 8c → 25.<<
Input @
"Switch Solve, MDRealOnly, Pure Solve";
MDRealOnly@Solve@82 × 25 + 2 b 100<, 8b<DD
88b → 25<<
Answer: The swimming pool should have a dimensions 25x25 for the biggest area
18.
Which number added to its square gives the smallest total amount?
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Note: Let the x will be te searched number.
Input @
Clear@f, xD;
f@x_D = x + x2
x + x2
Note: We will use a find minimum function to solve that problem
Input @
FindMinimum@f@xD, 8x<D
8−0.25, 8x → −0.5<<
Answer: This number is -0.5
Physics problems
Open ê Close
Note: This problems cannot be solved by student of technical school without using
Mathematica program
Relativistic mass
19.
In relativistic physics mass of object isn't constant. It is variable from velocity of
this mass. Plot the funcion describing mass relation from velocity for starting
mass(m0 ).Conclude the result.
m0 =500[g]
c=3 * 10 8 [m/s]
Note: Getting started we need to declare starting mass(m 0 ) and also for input
simplification light's speed(c)
Input @
clear@c, v, m0 D; m0 = 500; c = 3 ∗ 108 ;
Note: Inputing the funcion (f(v))
f@v_D =
Input @
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m0
2
1 − NBI vc M F
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500
1 − 1.11111 × 10−17 v2
Note: As we see program itself put our declared variables to funcion. Now the last thing
left to do is ploting the funcion in range from 0 to c.
Input @
MDPlot@8f@vD<, 8v, 0, c<,
PlotStyle → 88Red, [email protected]<<, PlotRange → 80, 1400<D
y
1400
1200
1000
800
600
400
200
0
x
5.0 µ 107 1.0 µ 108 1.5 µ 108 2.0 µ 108 2.5 µ 108 3.0 µ 108
Answer: At the funcion graph we see mass relation to velocity. When mass velocity is
close to light speed relativistic mass is multiple times higher then stoic mass of object.
20.
Find the object velocity at which relativistic mass ratio to stoic mass equals
n= 4/3.
c=3*10^8[m/s]
Note: First we need to declare our variables: mass ratio,lightspeed
Input @
clear@m0 , m, v, c, nD; n = 4 ê 3; c = 3 ∗ 108 ;
Note: Second we need to input our formula.
Input @
v = f@n_D = NBc ∗
1−
1
n2
F
1.98431 × 108
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Answer: Velocity at which mass ration equal 4/3 is 1.98431*108 [m/s] and is lower than
lightspeed which was easy to guess.
Waves
21.
Equation describing source vibration is y=0.04*sin(600pi*t). Vibrations are in
elastic envirioment. Find period of this vibration, kinematic wave equation and
describe deviation from balance point which is in distance of 75cm from source
after the time of 0,01 s from starting moment.Wave speed v=300m/s.Plot the
first equation
Note: First we need to find period from pulsation formula w=
2p
.
T
After formula transformation we can find the period. T=
2p
w
Clear@T, ωD;
ω = 600 π;
Input @
2π
T = f@ωD =
;
ω
Note: We have found the funcion period and it equals 1/300[s]. Kinematic wave formula
t x
is described by y=Asin(2p( - Å )). We can also evaluate lenght of the wave from
T l
l
velocity formula v= .
T
A = 0.04 ;
t = 0.01;
Input @ v = 300;
x = 0.75;
λ = f@v_D = T ∗ v;
Note: Now we have all the data so we can evaluate wave deviation.
Input @
t
EvaluateBSinB2 π ∗
T
−
x
λ
FF
1.
Input @
y = A ∗ SinB2 π ∗
t
T
−
x
λ
F
0.04
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Note: Last thing to do is to plot the equation y=0.04sin(600pt)
Input @
Clear@tD;
f@t_D = 0.04 ∗ Sin@600 π ∗ tD;
Input @
Plot@8f@tD<, 8t, 0, 0.05<, AxesLabel → 8 "t", "f@tD" <D
f@tD
0.04
0.02
0.01
0.02
0.03
0.04
0.05
t
-0.02
-0.04
Answer: Wave deviation equals 0.04 which is amplitude of this wave.
Trigonometry
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Print
Solved problems
Open ê Close
1.
A boy 1,5 m tall is standing 15 m away from an inflatable clown. He is staring up
at a 45° angle. How tall is the clown?
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Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
angle = 45;
oppositeLeg = 15;
adjacentLeg = oppositeLeg ê Tan@angle ê 180 ∗ πD êê N
15.
We calculate distance BC by tan.
Input @
15 + 1.5
16.5
The height of the clown is sum distance of BC and the height of the boy.
Answer: The clown is 16,5 m tall.
2.
A little kid is flying a balloon. The string of the balloon makes an angle of 30
degrees with the ground. If the height of the balloon is 12 m, find the length of
the string that the boy has used.
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Æ c ê Sin@gD ;
a = 12;
α = 30;
Input @ β = 90;
b=
a
SinA α∗π
E
180
∗ SinB
β∗π
180
F êê N
24.
To find the lenght of the string, we use the law of sines, because we know two angles
and the height of the balloon.
Answer: The lenght of the string is 24 m.
3.
Two ships, A and B leave Mersey Docks (C) at the same time. Ship A travels at
a bearing of 120° and ship B travels at a bearing o f 100°. After 1 hour ship A has
travelled 25 km and angle CBA is 130°. Find the spe ed of ship B.
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120°-100° = 20° > the angle ACB is 20°
180° - 130° -20° = 30° > the angle CAB is 30°
We use the law of sines.
Æ c ê Sin@gD ;
b = 25;
β = 130;
Input @ α = 30;
a=
b
β∗π
SinA 180 E
∗ SinB
α∗π
180
F êê N
16.3176
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Answer: The speed of ship B was 16,32 km/h.
4.
The sides of triangle ABC are 39 cm, 42 cm, 45 cm. The second longest height
of triangle is 36 cm. What is the shortest height?
We know: a = 42 cm, b = 39 cm, c = 45 cm, h = 36 cm
In order to find the angle b, we use the law of sines.
Æ Sin@gD ê c ;
δ = 90;
c = 45;
Input @ h = 36;
β = ArcSinBh ∗
SinA δ∗π
E
180
c
F∗
180
π
êê N
53.1301
Now we use the law of sines to measure x.
Æ a ê Sin@aD ;
a = 42;
γ = 90;
Input @ β = 53.1301;
x=
a
γ∗π
SinA 180 E
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∗ SinB
β∗π
180
F êê N
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33.6
Answer: The shortest height of triangle is 33,6 cm.
5.
Two towers face each other separated by a distance of 30 m. As seen from the
top of the first tower, the angle of depression of the second tower´s base is 60°
and that of the top is 30°. What is the height of t he second tower?
First, we find x and y by sine. The height of the second tower is x+y.
Input @
angle = 30;
adjacentLeg = 30;
oppositeLeg = adjacentLeg ∗ Tan@angle ê 180 ∗ πD êê N
17.3205
Input @
x = 17.32050807568877
17.3205
Input @
angle = 60;
adjacentLeg = 30;
oppositeLeg = adjacentLeg ∗ Tan@angle ê 180 ∗ πD êê N
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51.9615
Input @
y = 51.96152422706631
51.9615
Input @
h=x+y
69.282
Answer: The height of the second tower is about 69 m.
6.
A plane is flying above a place A. Danny can see the plane from a place B (B is
2 400 m away from A) at an angle of elevation of 54°20'. What is the height of
the plane above the Earth?
Angle ABC is 54°20'. We can calculate AC by tan.
Input @
angle = 54.33;
adjacentLeg = 2400;
oppositeLeg = adjacentLeg ∗ Tan@angle ê 180 ∗ πD êê N
3343.65
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Answer: The plane is flying 3343,65 m above ground.
7.
The walker is standing 150 m from Orava castle. He can see the top of the
castle at 37° What is the height of the castle?
Æ a ê Sin@aD ;
AB = 150;
γ = 53;
Input @ β = 37;
b=
AB
γ∗π
SinA 180 E
∗ SinB
β∗π
180
F êê N
113.033
We know AB and all angles, so we use the law of sines to calculate the height of the
castle.
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Answer: The height of the castle is 113,033 m.
8.
Input @
The first person is standing on the bank of the Orava dam and is 900 m far from
the second person, who is standing on the other side of the dam. The third
person can see the first one from a glider at 70° a nd the second at 60°.
Calculate the height of the glider.
20 + 30
50
This is the third angle in triangle 12A
Æ a ê Sin@aD ;
c = 900;
γ = 50;
Input @ β = 60;
b=
c
γ∗π
SinA 180 E
∗ SinB
β∗π
180
F êê N
1017.46
We need to know distance 1A. We can use the law of sines.
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Æ c ê Sin@gD ;
b = 1017.4642873624224`;
γ = 90;
α = 70;
a=
b
γ∗π
SinA 180 E
∗ SinB
α∗π
180
F êê N
956.104
We use the law of sines to calculate the height of the glider.
Answer: The height of the glider is 956,104 m.
9.
The statue in Klin ,,Rio de Klin" is 9,5 m tall. Two people are standing on the
right and left side of the statue. The person who is standing on the right can see
the top of the statue at 65°. The distance of perso n on the left from the statue is
1,5 times smaller than the distance of person on the right. At what angle can the
person on the left see the top of the statue?
Æ a ê Sin@aD ;
v = 9.5;
β = 65;
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65;
δ = 25;
SB =
v
β∗π
SinA 180 E
∗ SinB
δ∗π
180
F êê N
4.42992
We calculate distance of person on the right side from the statue by sin.
Input @
4.42992 ê 1.5
2.95328
The distance of person on the left is 1.5 times smaller, so it is 2,95328 m.
We use cosine function to find angle a.
Input @
oppositeLeg = 9.5;
adjacentLeg = 2.95328;
angle = ArcTan@oppositeLeg ê adjacentLegD ∗ 180 ê π êê N
72.731
Answer: The person on the left can see the top of the statue at 72,731°.
10.
On shore of the river there is a building. From one window of this building you
can see the point on the other side of the river under 25,3°. From other window
which is about 12 metres higher you can see this point under 37,5°. How wide is
that river?
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At first we have to calculate the size of angle AW2P.
Input @
angle1 = 37.5;
angle2 = 90;
AW2P = 180 − angle1 − angle2
52.5
Now we can find out W1P by the law of sines, angle W1PW2 = 37.5°-25.3°=12.2°
Æ a ê Sin@aD ;
b = 12;
β = 12.2;
Input @ γ = 52.5`;
c=
b
β∗π
SinA 180 E
∗ SinB
γ∗π
180
F êê N
45.0503
W1P = 45.0503 m. Lets calculate the size of angle AW1P and after that we can use the
law of sines again to find out the width of the river.
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Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
angle1 = 25.3;
angle2 = 90;
AW1P = 180 − angle1 − angle2
64.7
Æ b ê Sin@ bD ;
W1P = 45.05027447408226`;
α = 90;
Input @ AW1P = 64.69999999999999`;
c=
W1P
E
SinA α∗π
180
∗ SinB
AW1P ∗ π
180
F êê N
40.7292
Answer: The width of the river is 40,7292 m.
11.
The area lighted by a circular lamp has a conic shape. The angle of top of this
cone is 36°. The lamp is hanging on a mast which is 25 m high. An angle
between an axis of the mast and an axis of the top angle is 42°. How long is
the lighted stretch of the road?
At first we can calculate the distance between A and C by tangens function in right
triangle ADC. Angle ADC is 42° + 18° = 60°
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angle = 60;
adjacentLeg = 25;
oppositeLeg = adjacentLeg ∗ Tan@angle ê 180 ∗ πD êê N
43.3013
AC = 43,3013 m. Now we can find out AB by tangens function, too. Angle ADB is 42° 18° = 24°.
Input @
angle = 24;
adjacentLeg = 25;
oppositeLeg = adjacentLeg ∗ Tan@angle ê 180 ∗ πD êê N
11.1307
AB = 11,1307 m.
Input @
BD = 43.30127018922193` − 11.130717132713402`
32.1706
Answer: The length of the lighted strech is 32,1706 m.
12.
Target C is watched from two gunnery positions A and B which are 296 m from
each other. We know that angle BAC is 52 ° and ABC is 44 °. How far is A from
C?
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g=180°- (52°+ 44°)
Input @
angle1 = 52;
angle2 = 44;
γ = 180 − angle1 − angle2
84
Æ c ê Sin@gD ;
c = 296;
γ = 84;
Input @ β = 44;
b=
c
γ∗π
SinA 180 E
∗ SinB
β∗π
180
F êê N
206.751
Answer: The distance between A and C is 206,751 m.
13.
In the triangle ABC the ratio of its sides is a:b:c = 3 :1:2 and the angle ABC is
60°. Solve: sin(ABC) + sin (BCA) = ?
Using Pythagoras´ Theorem we can show that ABC is a right triangle.
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firstLeg = 1;
Input @
secondLeg =
hypotenuse =
3;
firstLeg2 + secondLeg2 êê N
2.
hypotenuse = 2;
Input @
adjacentLeg =
3;
α = ArcCos@adjacentLeg ê hypotenuseD ∗ 180 ê π êê N
30.
Input @
β = 60
γ = 90
60
90
Input @
SinB
Input @
1+
β∗π
180
3
2
F + SinB
γ∗π
180
F
êê N
1.86603
Answer: sin(ABC) + sin (BCA) is 1,86603.
14.
The army convoy is marching along a straight road. We are in the place A. The
distance between the begining of the army convoy (B) and A is 14 350 m. From
the end of the convoy (C) to A 13 840 m. The angle BAC is 51°. How long is the
army convoy?
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We know: b=13840
c=14350
a=51°
We calculate the length of the army convoy by the law of cosines.
b = 13 840;
c = 14 350;
Input @ α = 51;
a=
b2 + c2 − 2 ∗ b ∗ c ∗ CosBα ∗
π
180
F êê N
12 144.8
The army convoy is long 12 144,8 m.
15.
Two ships are floating from the island ,,Slanica". When the first ship crosses 30
metres, it can see the island ,,Slanica" at angle of elevation of 38°. After 20
metres, the second ship can see the island ,,Slanica" at angle of elevation of
46°. Calculate the distance of two ships from each other.
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angle1 = 38;
Input @
angle2 = 46;
γ = 180 − angle1 − angle2
96
We know two angles - 38° and 46°, so we can easily find the third angle - 180° - ( 38°
+ 46° ) = 96°
a = 20;
b = 30;
Input @ γ = 96;
c=
a2 + b2 − 2 ∗ a ∗ b ∗ CosBγ ∗
π
180
F êê N
37.7549
We calculate the distance of two ships by the law of cosines, because we know two
sides and an angle.
Answer: The distance of two ships is 37,76 metres.
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We can see the top of tower from a place A at an angle of elevation of 37°.
When we get 50 metres nearer to the bottom of the tower - B, the angle of
elevation will be 58°. How high is the tower?
Input @
β = 58;
δ = 180 − β
122
α = 37;
Input @ δ = 122;
γ = 180 − Hα + δL
21
We know all angles in the triangle ABV.
Distance AV we can calculate by the law of sines
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Æ a ê Sin@aD ;
v = 50;
γ = 21;
Input @ δ = 122;
AV =
v
γ∗π
SinA 180 E
∗ SinB
δ∗π
180
F êê N
118.321
We calculate the height of the tower by sine function
Input @
angle = 37;
hypotenuse = 118.321;
oppositeLeg = Sin@angle ê 180 ∗ πD ∗ hypotenuse êê N
71.2074
Answer: The tower is 71,2074 metres high.
17.
Two trains are coming out of a station following two straight lines which make
the angle of 156°. The speed of the first train is 13 m/s, the speed of the second
one is 14,5 m/s. How far from each other will these two trains be in 5,5 minutes?
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Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
5.5 ∗ 60
330.
Now we claculate how many seconds there are in 5,5 minutes.
Input @
13 ∗ 330
4290
Input @
14.5 ∗ 330
4785.
We calculate distance 1S and 2S by formula: s(distance)=v(speed)·t(time).
b = 4785;
c = 4290;
Input @ α = 156;
a=
b2 + c2 − 2 ∗ b ∗ c ∗ CosBα ∗
π
180
F êê N
8877.29
Distance 12 we can calculate by the law of cosines.
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Answer: After 5,5 minutes two trains will be 8877,29 metres apart.
18.
The plane is flying 3000 m above the ground. At first, a person standing on the
ground can see this plane at an angle of 25° , afte r a while, at an angle of 50°.
How far did the plane fly during that time?
At first we have to calculate AL1 by cosine function in the right triangle ABL1. Angle
AL1B is 90° - 25° = 75°
Input @
angle = 75;
adjacentLeg = 3000;
hypotenuse = adjacentLeg ê Cos@angle ê 180 ∗ πD êê N
11 591.1
AL1 = 11591.1 m. We can use the law of sines to find out L2L1. Angle AL2L1 = 180° 50° = 130° . And we have to calculate the size of a ngle L2AL1.
Input @
angle1 = 130;
angle2 = 25;
L2AL1 = 180 − angle1 − angle2
25
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Æ b ê Sin@ bD ;
c = 11591.109915468822`;
γ = 130;
α = 25;
x=
c
γ∗π
SinA 180 E
∗ SinB
α∗π
180
F êê N
6394.69
Answer: The plane flew over 6394,69 m during that time.
19.
From two places which are 3100 m far from each other two people can see a
cloud in the sky at angles of 65° and 79° . How hig h is the cloud?
In the picture you can see that angle MNA is 79° , too and angle NMA is 65° . Now lets
calculate the third angle in the triangle MNA.
Input @
angle1 = 79;
angle2 = 65;
MAN = 180 − angle1 − angle2
36
Now we can find out NA by the law of sines.
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Æ b ê Sin@ bD ;
c = 3100;
γ = 36;
α = 65;
a=
c
γ∗π
SinA 180 E
∗ SinB
α∗π
180
F êê N
4779.9
NA = 4779.9 m. Lets find out how high the cloud is. We can use sine function in the
right triangle BNA.
Input @
angle = 79;
hypotenuse = 4779.899000282544`;
adjacentLeg = Sin@angle ê 180 ∗ πD ∗ hypotenuse êê N
4692.08
Answer: The cloud is 4692,08 m above the ground.
20.
Tourists were following a map, from place A to B they went 2 km straight north,
from B to C at a bearing of 53° 13 km and from C to D at a bearing of 90° 4 km.
What is the direct distance from A to D?
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We know a=53°, b=90°, −BEC=d=90°. So −BCE=g must be 37°.
Æ c ê Sin@gD ;
BC = 13;
δ = 90;
Input @ γ = 37;
BE =
BC
SinA δ∗π
E
180
∗ SinB
γ∗π
180
F êê N
7.8236
Æ c ê Sin@gD ;
BC = 13;
α = 53;
Input @ δ = 90;
EC =
BC
SinA δ∗π
E
180
∗ SinB
α∗π
180
F êê N
10.3823
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We used the law of sines to find BE, EC. In order to find AD, we use the right triangle
ADE in which AE=AB+BE=9,236 and ED=EC+CD=14,3823.
firstLeg = 9.236;
secondLeg = 14.3823;
Input @
hypotenuse =
firstLeg2 + secondLeg2 êê N
17.0925
Answer: The distance from A to D is 17,0925 km.
21.
Lomnický š tít is 2632 m a. s. l.. There is a cable railroad from Skalnaté pleso
(L) to Lomnický štít (V), see the picture. What is the altitude of Skalnaté pleso?
How long does it take to get from L to V, if the speed of the cable railroad is 4,1
m/s?
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We can find LV by the law of sines.
Æ b ê Sin@ bD ;
XL = 1653;
γ = 70;
Input @
α = 90;
XL
α∗π
LV =
∗ SinB
F êê N
γ∗π
180
SinA 180 E
1759.09
hypotenuse = 1759.0858579026826`;
givenLeg = 1653;
Input @
leg =
hypotenuse2 − givenLeg2 êê N
601.643
Skalnate pleso is 2632 - 601.643= 2030,357 m a.s.l.
t (time) = s (distance) / v (speed)
LV = 1759.09;
Input @ v = 4.1;
t = LV ê v
429.046
Input @
429.04634146341465` ê 60
7.15077
Answer: The time is about 7,15 minutes and Skalnate pleso is 2030,357 m a.s.l.
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Probability
Open ê Close
Print
Calculation Theorems
Open ê Close
Definitions
Sample Spaces
A sample space is the set of all possible outcomes.
Example 1. Consider the experiment of flipping two coins. It is possible to get 0 heads, 1
head, or 2 heads. Thus, the sample space could be the
set {0, 1, 2}. Another way to look at it is flip { HH, HT, TH, TT }.
In the table the 1st flip is in the horizontal and the 2nd flip in the vertical line and the four
outcomes are in black.
Ñ H T
H HH HT
T HT TT
The second way is better because each event is as equally likely to occur as any other.
A classical probability
The probability of an event occurring is the number of cases favorable for the event n (E),
over the number of total outcomes possible n (S). This is only true when the events are
equally likely.
P(E) =
n HEL
n HSL
Empirical Probability
Empirical probability is based on observation. The empirical probability of an event is the
relative frequency of a frequency distribution based upon observation.
P(E) =
f
Å
n
Example 2. This time you are rolling two dice. The sums are { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12 }. However, each of these aren't equally likely. The only way you can get a sum 2 is to
roll a 1 on both dice, but you can get a sum of 4 by rolling a 1-3, 2-2, or 3-1. The following
table illustrates a better sample space for the sum obtain when rolling two dice. In the
lowest horizontal row are the points of the first dice and in the most left column are the
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points of the second dice. The sum of the two dice are counted in the
cells.
6
5
4
3
2
1
results
7
6
5
4
3
2
1
8
7
6
5
4
3
2
9 10 11 12
8 9 10 11
7 8 9 10
6 7 8 9
5 6 7 8
4 5 6 7
3 4 5 6
Now it's easy to find for example the propability to get sum 4.
3
1
P(" the sum is 4") =
=
.
36
12
In the next table we have a probability distribution:
Sum Relative frequence
2
1
36
3
2
36
4
3
36
5
4
36
6
5
36
7
6
36
8
5
36
4
9
36
10
3
36
11
2
36
12
1
36
Total
36
36
= 1
Example 3. The propability to get sum between 4 and 8 is
3
4
5
6
5
23
+
+
+
+
=
.
36
36
36
36
36
36
In the next table you find the running total of the relative frequencies or the cumulative
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frequency divided by the total frequency.
Sum Cumulative Relative Frequency
1
2
36
3
3
36
4
6
36
5
10
36
6
15
36
7
21
8
26
9
30
36
10
33
11
35
36
12
36
36
36
36
36
Again, find the propability to get sum between 4 and 8.
1. Complementation theorem
For the complementary event A of event A exists a theorem
P(A) = P(A doesn't happen) = 1-P(A)
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Example 4.
The odds for a serial producted single-use item being malfunctional is 1.5% What is the
probability for having at least one malfunctioning product in a set of hundred?
Because P(product is malfunctional) = 1.5%, is
P(product is functional) = 98.5%
Event "at least one is malfunctional" considers the possibility that there are 1 or 2 or 3 or
... or 100 malfunctioning products. The complementary event is "there are no
malfunctional products" ie. "all products are functional".
P(at least one is malfunctional) = 1 - P(all products are functional) = 1 100
0.985
= 0.7793... º 78%
2. Multiplication theorem
If events A and B do not affect each other in any way, being independant, there is a
possibility that the events A and B take place
P(A and B) = P(A)×P(B)
If the events A and B are dependant of each other the multiplication theorem is valid in the
form
P(first A, then B) = P(A)×P(B, when A has already taken place)
Example 5.
A die is thrown three times. The results are independant of each other. The odds that
the result is a series of 1, 2 and 3 are thus
1 1 1
1
º 0.5%
P(1,2,3) = Å × Å × Å =
6 6 6 216
Three tags are taken at random from a box in which contains the number tags 1, 2, 3, 4,
5 and 6. The results are not independant of each other. The probability for the result of
three consequent picks being the series 1, 2 and 3 is
1 1 1
1
P(1,2,3) = Å × Å × Å =
º 0.8%
6 5 4 120
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1.
Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
The pincode of a bank card is four-digit. What are the odds that a person who
only remembers the last digit will succeed at entering the code on his first try?
P HAL =
1
×
10
1
10
×
1
10
×1
1
1000
2.
a)
Four cards are taken at random from a deck of cards. Count the probability for
a) All four being aces
b) Not a single one being an ace
c) Atleast one being an ace
A = "All are aces"
P HAL =
Input @
=
4
×
52
1
270 725
3
51
×
2
50
×
1
49
êê N
3.69379 × 10−6
b)
B = "Not a single one is an ace"
Input @
P HBL =
48
38 916
êê N
54 145
52
×
47
51
×
46
50
×
45
49
0.718737
c)
B = "Atleast one is an ace"
P(B) = 1 - P(B)
38 916
¯
P HBL = 1 −
54 145
Input @
15 229
êê N
54 145
0.281263
Answer:
1
a)
º 3.69379 × 10−6 º 0,0000037
270 725
b)
38 916
º 0.718737º 0,719
54 145
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c)
15 229
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
º 0.281263 º 0,281
54 145
3.
Two balls are taken at random from a box which contains 3 red, 4 blue and 5
black balls. What is the probability for them being the same color?
P("both are red") + P("both are blue") + P("both are black) =
Input @
Input @
3
2
12 11
19
66
+
4
3
12 11
+
5
4
12 11
êê N
0.287879
Answer:
19
66
º 0.287879 º 0,288
3. Tree model
If the phenomenom being studied is divided into separate events, which together cover the
whole phenomenom, it is said that the sum of these probabilities = 1 (=100%). In the case
of seperate events it is possible to talk about the addition theorem.
The overall situation is often depictable with a tree model, where different events are
recorded with their respective probabilities. A tree model is read from top to bottom, so that
one picks the branch that describes a series of events, and then multiply all the different
odds of that branch. In the end, the probabilities of different branches are then summed.
Tree Diagram
A graphical device used to list all possibilities of a sequence of events in a systematic way.
Example 6.
There are three production lines in a factory, A, B and C. Of the total production 40%
goes through line A, 25% through line B and the rest through line C. 3% of the products
of line A, 1% of the products of line B and 2% of the products of line C are
malfunctional. A random product is picked from a large batch. What are the odds that
the product in question is
malfunctional?
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It is possible to calculate the answer directly from the model.
P(the product is malfunctional) = 0.40 × 0.03+ 0.25 × 0.02 + 0.35
0.0205 º 2%
0.01 =
In practice we have come along the addition theorem as a part of the tree model's
solution. When a problem includes different choices and options, then
The consequent events "first A, then B" point to
multiplication
The parallel events "A or B" point to addition
4.
It is expected that the following probabilities take place:
1º If it doesn't rain on any chosen day , the probability for rain the following day
1
is Å
3
1
2º If it rains on any chosen day, the probability for sunshine the following day is Å
2
What is the probability for rain the day after tomorrow, if it rains today?
Tree model! A= "Rain tomorrow, when it rains
today"
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Input @
P HAL =
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
1
2
×
1
3
+
1
2
×
1
2
5
12
4. Addition theorem
If the events A and B are independant, ie. when the other takes place the other cannot,
there is a probability that the result of the random phenomenon is "A or B".
P(A or B) = P(A) + P(B)
Example 7.
What is the probability for a two-figure number being divisible by
a) two b) three c) six d) two or three?
Two-figure numbers:
Numbers divisible by two:
Numbers divisible by three:
Numbers divisible by six:
a) P(divisible by 2) =
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10, 11, 12, ..., 99;
190 altogether
10, 12, ..., 98; 45 altogether
12, 15, ..., 99; 30 altogether
12, 18, ..., 96; 15 altogether
45 1
= Å = 50%
90 2
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30 1
= Å º 33.3%
90 3
15 1
c) P(divisible by 6) =
= Å º 16.7%
90 6
d) Because numbers divisible by six are divisible by both two and three they are
accountable in the amount of numbers divisible by two as well as three. In other words,
they are involved twice in the addition 45+30. Thus, there are 45 + 30 - 15 numbers
divisible by two or three.
b) P(divisible by 3) =
P(divisible by 2 or 3) =
45 + 30 - 15
90
=
60
2
= Å = 66.7%
90 3
Example 8.
A sweets factory produces its "mix of the best"-assortment, of which 45% of the sweets
contain additive A and 32% additive B. Both additives are found in 15% of the sweets. A
candy is picked from a batch at random. What are the odds that the candy contains a)
atleeast one of the two additives b) neither of the additives?
Now we cannot use figures the same way we did in example 6.
Let's make it easier to grasp the situation by tabulating the given information. Since both
additives can be found in 15% of the sweets, the percentage of sweets containing only
additive A is 45-15 = 30 and only additive B 32-15 = 17.
Additive B
Positive
Negative
Additive A
Positive
Negative
15
32-15=17
45-15=30
a) P(atleast one additive) = 15% + 17% + 30% = 62%
b) In the remaining section, the cell corresponding the lower right corner of the table, no
additives have been used.
P(neither of the additives)
= 100% - (15% + 17% + 30%) = 38% of the sweets.
In the case of examples 7 and 8 the given events were not independant, but there was the
possibility of them taking place at the same time. In this case, like presented in the
examples, the elimination of the events' redundancy is required while calculating the
probability for 'A or B'.
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5. Probability with a certain condition
It is often on purpose that we examine event A, in relation with a random event, in a
smaller range of correspondance than the original population.
Example 9.
Let's think of the situation in example 6, where we calculated the probability for a
product being malfunctional and we got the result
P(the product is malfunctional) = 0.205
If we are especially interested in the probability for the malfunctional product having
been manufactured on production line A, it is natural to think of the group formed by the
malfunctional products as a completely new domain.
The probability for a product being from line A while acknowledging that the product is
malfunctional is found by examining the share of products from line A in comparison
with the number of malfunctional products from all production lines.
P(the malfunctional product is from production line A) =
0.40 µ 0.03
= 0.585... º 59%
0.0205
In these type of situations we speak of conditional probability. We will not go further into
the subject or using conditional probability on a wider scale.
5.
In a class of 35 all students study either French or German. 32 students study
French and 29 study German. How many study
a) both
b) only French
c) only German?
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French = grey
German = magenta
Answer a) 26 b) 6 c) 3
6.
Out of the two parking lots of a small store, both are taken 40 minutes an hour.
32 minutes of the time both lots are taken simultaneously. Two drivers arrive at
the store at the same time. What are the odds that these two will find parking
space immediately?
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Answer : P H"Both get a lot"L =
1 − P H"atleast one of the lots is taken"L =
Input @
8 + 32 + 8
12
1
1−
=
=
60
60
5
7.
What are the odds for a randomly chosen natural number to be divisible by
a) the numbers 4, 5 and 6?
b) the numbers 4, 5 or 6?
Every fourth natural number is divisible by 4.
Every fifth number is divisible by 5 and every sixth by 6.
Every twelth number is divisible by 4 and 6.
Every twentieth number is divisible by 4 and 5.
Every thirtieth number is divisible by 5 and 6.
Every sixtieth number is divisible by 4, 5 and 6.
Input @
LCM@4, 5, 6D
60
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Input @
1
4
+
1
5
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
+
1
6
−
1
12
+
1
20
+
1
30
9
20
Answer a)
1
1 1 1
1
1 1
9
b) Å + Å + Å +
=
+
60
4 5 6
12 20 30 20
Geometrical probability
Open ê Close
If the sample space is possible to be viewed as a geometrical groups such as line segment
(= length measure), sector (=angle or area measure), plane (= area measure) or space
area (= volume measure), the probability for event A is P(A)=
The geometrical measure of A
.
The measure of whole sample space
Example 1. Straight is drawn from point (1,3) to random direction.
What is the probability that straight intersects the positive x-axis?
Clear@f, xD;
Input @
f@x_D = 3 x
g@x_D = 3
3x
3
Input @
MDPlot@8f@xD, g@xD<, 8x , −1 , 3 < D
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"y"
8
6
4
Input @
b
b
2
a
b
-1
1
2
3
"x"
-2
Calculate angle a = 180°- b. Angle tan b = k (slope). The probability for
a
the event is the ratio P(A) =
.
180
Input @
Æ Pure NSolve
;
Clear@xD;
Input @
MDSelectSolution@MDRealOnly@NSolve@
8 Tan@β DegreeD == 3<, 8 β < DD, 1D
88β → 71.5651<<
Input @
β = 71.5651
71.5651
Input @
NumberForm@β, 4D
+71.57
Input @
α = 180 − β
Input @
108.43494882292201
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Input @
p HAL =
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
108.43
180
0.602389
Answer: 0.60
Example 2. The total length of the Channel Tunnel is 50 kilometers and the
underwater part is 38 km. A train which is 400 meters long stops at random place in this
railway between Calais and Dover. What is the probability that the train is
a) entirely
b) entirely or partly under water?
http://fi.wikipedia.org/wiki/Tiedosto:Course_Channeltunnel_en.png
Solution: Focus on the front of the locomotive and find the favorable distance.
a) The train is entirely under water if the front of the locomotive is not longer than 0,400
km from the rear end of the underwater tunnel.
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Input @
P HAL =
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
38 − 0.400
50
0.752
b) The train is entirely or partly under water if the front of the locomotive is inside the
underwater tunnel or not more than 0,400 km past the end of the tunnel.
Input @
P HBL =
38 + 0.400
50
0.768
Answer a) 0.752 b) 0.768
Example 3. The coefficients a and b of the straight ax - by = 0 are picked randomly from
1
between [0,1]. What is the probability that the slope of the straight is § Å ?
3
Solution:
a
a
The equation for the straight can be written to a form y = Å x, where the slope Å can be
b
b
a
1
taken. So § Å , when b ¥ 3 a. The shadowed area at the picture corresponds that. The
b
3
solution can be calculated by calculating the ratio of the areas of the triangle and the
square.
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Input @
1 × 13
Input @
P HAL =
2
1×1
1
6
1
Answer: Å
6
Example 4. The radius of the dartboard's bulsseye is 2 cm, score 9 is 4cm etc. The
perimeter is 1cm wide. The dart hits random score. Calculate the expected value.
Solution: Calculate (!!) each score's area of the circle divided by the area of the whole
board. At first calculate the area of the score "10", which is 4p.
n
So the area of the score "9" is p 4 - p 2 = 12p etc. Expected value = ‚ xi ÿpi .
2
2
i=1
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Input @
Distribution table
X
p
10
π 22
π 212
π 42 −π 22
π 212
π 62 −π 42
π 212
π 82 −π 62
π 212
9
8
7
Input @
6
π 102 − π 82
π 212
5
π 122 −π 102
π 212
4
π 142 −π 122
π 212
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3
π 162 −π 142
π 212
2
π 182 −π 162
π 212
1
π 202 −π 182
π 212
0
π 212 −π 202
π 212
Σ
1
Distribution = ::10,
:8,
Input @
Input @
20
4
441
>, :9,
>, :6,
63
52
4
4
147
>,
>,
49
20
:5,
>, :4,
>, :3,
>,
441
441
147
68
76
41
:2,
>, :1,
>, :0,
>>
441
441
441
4
4
::10,
>, :9,
>,
441
147
20
4
4
:8,
>, :7,
>, :6,
>,
441
63
49
44
52
20
:5,
>, :4,
>, :3,
>,
441
441
147
68
76
41
:2,
>, :1,
>, :0,
>>
441
441
441
Expected value =
4
4
20
10 ×
+ 9×
+8×
+
441
147
441
Site: www.deltasoft.at
441
44
>, :7,
4
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7×
Input @
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
4
63
20
+ 6×
4
+5×
44
+ 4×
52
+
49
441
441
68
76
41
3×
+ 2×
+1×
+ 0×
147
441
441
441
220
êê N
63
3.49206
Answer:
8.
Expected value º 3 .5
Solved Problem:
Britney and David arrive independantly at the same bar between 19:00 and
20:00 in the evenings, staying in the bar for 15 minutes each time. What's the
probability for the two being in the bar at the same time even for a moment?
Solution: X = "Britney's time of arrival" and Y = "David's time of arrival".
The two will meet if X - Y § 15 and Y - X § 15 meaning Y ¥ X - 15 and Y § X + 15. We
draw the lines Y = X - 15 and Y = X + 15. The favorable zone is the area lined out by the
two lines.
All possibilities are depicted by a 60x60 square.
Input @
MDPlot@8x − 15, x + 15, 60<, 8x , 0 , 60 <,
PlotRange → 88 0, 60 < , 80, 60<<, Filling → 81 → 82<<D
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Input @
Input @
602 − 2 ×
45×45
2
602
7
16
Answer: P("meet") =
7
16
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Binomial probability
Open ê Close
Permutation
An arrangement of objects in a specific order.
In your Calculator, the button nPr.
If there are n elements in a set, the elements can be arranged
in n! different orders. The arrangements are called
permutations.
From a set of n elements can be formed n*(n-1)*(n-2)* ...
*(k-1) = n [nPr] permutations of k elements.
Example 1. Seven brothers (of Jukola) can arrange
themselves in 7! = 5040 different orders.
Example 2. The brothers can elect a chairman, secretary and
a treasurer in 7*6*5 = 7 [nPr 3] = 210 ways.
Combination
A selection of objects without regard to order.
In your Calculator, the button nCr.
If there are n elements in a set, you can form subsets which
have k elements in
n
n!
=
= n [nCr] k ways.
k
Hn - kL! k!
In Mathematica
Site: www.deltasoft.at
n
k
= Binomial[n,k].
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These subsets are called combinations.
Example 3. The seven brothers can elect 3 people amongst
themselves to go to the store in
7
7!
=
= 7 [nCr] 3 = 35 ways.
3
H7 - 3L! 3!
Example 4. In a mycology course the students were taught to
idetify 78 different mushrooms, of which one of the students
only learned 49. What is the probability that he would identify
6 randomly selected mushrooms.
Solution:
78
= 256 851 595.different groups of 6
6
mushrooms.
49
There are
= 13 983 816. favorable cases.
6
The probability that all the mushrooms will be identified
correctly is
49
6
= 0,054
78
6
There are
Example 5. How big must a class be that it is favorable to bet
(with even odds) that at least two students have the same
birthday?
Solution:
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The number of students = n
P("At least two have the same birthday") =
1- P("all have different birthdays") =
1-
365 364 363
365 - Hn - 1L
.
.
.. …
. > 0,50
365 365 365
365
365 364 363
365 - Hn - 1L
.
.
.. …
< 0,50
365 365 365
365
Calculate the smallest value for n, so that
365 @nPrD n
365n
< 0,50.
Table the values:
n
365 @nPrD n
10
365 @nPrD µ 10
365n
365
20
0,589
25
0,421
24
0,462
23
0,493
22
0,52
Site: www.deltasoft.at
10
= 0,883
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Answer: there must be at least 23 students.
Binomial Experiment
A binomial experiment is an experiment which satisfies these
four conditions
⋅
A fixed number of trials
⋅
Each trial is independent of the others
⋅
There are only two outcomes
⋅
The probability of each outcome remains constant from
trial to trial.
These can be summarized as: An experiment with a fixed
number of independent trials, each of which can only have two
possible outcomes.
The fact that each trial is independent actually means that the
probabilities remain constant.
Examples of binomial experiments
⋅
Tossing a coin 100 times to see how many heads occur.
⋅
Asking 500 people if they watch news.
⋅
Rolling a die to see if a 6 appears.
Examples which aren't binomial experiments
⋅
Rolling a die until a 6 appears (not a fixed number of
trials)
⋅
Asking 20 people how old they are (not two outcomes)
⋅
Drawing 5 cards from a deck for a poker hand (done
without replacement, so not independent)
Binomial Probability Function
Example 6:
What is the probability of rolling exactly two sixes in 6 rolls of a
die?
There are five things you need to do to work a binomial story
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problem.
Define Success first. Success must be for a single trial.
Success = "Rolling a 6 on a single die"
Define the probability of success (p): p = 1/6
Find the probability of failure: q = 5/6
Define the number of trials: n = 6
Define the number of successes out of those trials: k = 2
Anytime a six appears, it is a success (denoted S) and
anytime something else appears, it is a failure (denoted F).
The ways you can get exactly 2 successes in 6 trials are given
below. The probability of each is written to the right of the way
it could occur. Because the trials are independent, the
probability of the event (all six dice) is the product of each
probability of each outcome (die)
1 FFFFSS 5/6 * 5/6 * 5/6 * 5/6 * 1/6 * 1/6
1 2 5 4
= Å
Å
6
6
2 FFFSFS 5/6 * 5/6 * 5/6 * 1/6 * 5/6 * 1/6
1 2 5 4
Å
= Å
6
6
3 FFFSSF 5/6 * 5/6 * 5/6 * 1/6 * 1/6 * 5/6
1 2 5 4
= Å
Å
6
6
4 FFSFFS 5/6 * 5/6 * 1/6 * 5/6 * 5/6 * 1/6
1 2 5 4
= Å
Å
6
6
5 FFSFSF 5/6 * 5/6 * 1/6 * 5/6 * 1/6 * 5/6
1 2 5 4
= Å
Å
6
6
6 FFSSFF 5/6 * 5/6 * 1/6 * 1/6 * 5/6 * 5/6
1 2 5 4
= Å
Å
6
6
7 FSFFFS 5/6 * 1/6 * 5/6 * 5/6 * 5/6 * 1/6
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1 2 5 4
= Å
Å
6
6
8 FSFFSF 5/6 * 1/6 * 5/6 * 5/6 * 1/6 * 5/6
1 2 5 4
Å
= Å
6
6
9 FSFSFF 5/6 * 1/6 * 5/6 * 1/6 * 5/6 * 5/6
1 2 5 4
Å
= Å
6
6
10 FSSFFF 5/6 * 1/6 * 1/6 * 5/6 * 5/6 * 5/6
1 2 5 4
= Å
Å
6
6
11 SFFFFS 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 1/6
1 2 5 4
Å
= Å
6
6
12 SFFFSF 1/6 * 5/6 * 5/6 * 5/6 * 1/6 * 5/6
1 2 5 4
Å
= Å
6
6
13 SFFSFF 1/6 * 5/6 * 5/6 * 1/6 * 5/6 * 5/6
1 2 5 4
= Å
Å
6
6
14 SFSFFF 1/6 * 5/6 * 1/6 * 5/6 * 5/6 * 5/6
1 2 5 4
= Å
Å
6
6
15 SSFFFF 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6
1 2 5 4
= Å
Å
6
6
Notice that each of the 15 probabilities are exactly the
1 2 5 4
same: Å
Å .
6
6
1
Also, note that the Å is the probability of success and you
6
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5
needed 2 successes. The Å is the probability of failure, and if
6
2 of the 6 trials were success, then 4 of the 6 must be failures.
Note that 2 is the value of x and 4 is the value of n-k.
Further note that there are fifteen ways this can occur. This is
the number of ways 2 successes can be occur in 6 trials
without repetition and order not being important, or a
combination of 6 things, 2 at a time.
The probability of getting exactly x success in n trials, with the
probability of success on a single trial being p is:
n
P(X=k) =
pk qn-k where q = 1-p (or P(X=k) = nCk
k
* pk * qn-k )
Example 7:
A coin is tossed 10 times. What is the probability that exactly 6
heads will occur.
Success = "A head is flipped on a single coin"
p = 0.5
q = 0.5
n = 10
x=6
10
P(X=6) =
o .56 o .510-6 (or 10C6 * 0.5^6 * 0.5^4) = 210 *
6
0.015625 * 0.0625 = 0.205078125
Mean, Variance, and Standard Deviation
The mean, variance, and standard deviation of a binomial
distribution are extremely easy to find.
m = np
d2 = npq
d=
npq
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Example 8.
Find the mean, variance, and standard deviation for the
number of sixes that appear when rolling 30 dice.
Success = "a six is rolled on a single die". p = 1/6, q = 5/6.
The mean is 30 * (1/6) = 5. The variance is 30 * (1/6) * (5/6) =
25/6. The standard deviation is the square root of the variance
= 2.041241452 (approx)
Example 9. When a coin is thrown five times. What are the
changes of getting second heads with the last throw?
The heads has to occur exactly one time in the 4 first throws.
P("heads occurs second time in the last
4
throw")=
1
1
Å
2
1
1
Å
2
31
1
Å =Å
2 8
1
Answer: The changes are Å .
8
In repeating the same random trial n times we receive the
probability for event A happening exactly k times
n k n- k
P(Ak ) =
p q , where the probability for a single
k
event A occuring is P(A) = p , and q = 1 - p.
9.
A) What is the probability for getting atleast two sixes on five dice rolls?
B) How many dice rolls are necessary for the probability for getting atleast one
six to rise over 90?
A) Solution:
By complementary event.
P("Atleast two sixes") = 1 - P("One six at most")
n = 5;
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p =
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
1
;
6
x = 0;
PDF@BinomialDistribution@n, pD, xD êê N
H∗ PH X = x L = ? ∗L
0.401878
Input @
p0 = 0.4018775720164609
0.401878
n = 5;
p =
1
;
6
Input @
x = 1;
PDF@BinomialDistribution@n, pD, xD êê N
H∗ PH X = x L = ? ∗L
0.401878
Input @
p1 = 0.4018775720164609
0.401878
Input @
1 − Hp0 + p1L
0.196245
Answer: 0.196 (=19.6)
B) the number of necessary throws=n
x1 = 1 ;
probability = 0.90 ;
Input @
MDSFindnBinomialDistributionPnp@ x1 ≤ X ≤ n,
8p, probability<D
H∗ find n that PHx1 ≤ X ≤ nL = probability ∗L
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PH1 § X § 13L º0.91
Probability
0.30
0.25
0.20
0.15
0.10
0.05
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Binomial Distribution
n
p
13
16
µ
σ
PH 1 ≤ X ≤ 13L
2.16667 1.34371 0.90653612101282076311
Answer: Atleast 13 throws.
10.
In the second round of the 2000 Finnish presidential elections Tarja Halonen
received 51,6% of the votes whereas Esko Aho received 48,4%. What is the
probability for a random group of five voters voting for Esko Aho?
n = 5;
p = 0.484 ;
Input @ x = 3 ;
MDSBinomialDistributionPnp@ x ≤ X , 8n, p<D
H∗ PH value ≤ X L = ? ∗L
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PH3 § XL º0.47
Probability
0.30
0.25
0.20
0.15
0.10
0.05
0
1
2
3
4
5
Binomial Distribution
n
p
µ
σ
PH 3 ≤ XL
5 0.484 2.42 1.11746 0.47002047370854400000
Answer: 0.47
11.
The probability for a certain seed type to sprout is 75%. How many seeds do you
have to plant in order to gain the probability for atleast one plant sprouting over
99,9%?
p = .75 ;
Input @
x1 = 1 ;
probability = 0.999 ;
MDSFindnBinomialDistributionPnp@ x1 ≤ X ≤ n,
8p, probability<D
H∗ find n that PHx1 ≤ X ≤ nL = probability ∗L
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PH1 § X § 5L º0.9990
Probability
0.4
0.3
0.2
0.1
0
1
2
3
4
5
Binomial Distribution
n
p
µ
σ
PH 1 ≤ X ≤ 5L
5 0.75 3.75 0.968246 0.99902343750000000000
or alternatively
p = 0.75 ;
x1 = 1 ;
probability = 0.999 ;
Input @ 8nstart = 1, nend = 6<;H∗ change nstart,nend ∗L
MDSFindnBinomialDistributionSteps@ x1 ≤ X ≤ n,
8p, probability<, 8nstart, nend<D
H∗ find n that PHx1 ≤ X ≤ nL = probability ∗L
n
1
2
3
4
5
6
PH x1 <= X <= n L Wanted P
0.75000000000000000000
0.93750000000000000000
0.98437500000000000000
0.99609375000000000000
0.99902343750000000000
0.99975585937500000000
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0.999
0.999
0.999
0.999
0.999
0.999
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PH x1 <= X <= n L
1.0
0.8
0.6
0.4
0.2
1
2
3
4
5
6
n
Answer: Atleast 5 seeds.
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Multinomial Probabilities
Open ê Close
Multinomial Probabilities
A multinomial experiment is an extended binomial probability. The difference is that in a
multinomial experiment, there are more than two possible outcomes. However, there are
still a fixed number of independent trials, and the probability of each outcome must remain
constant from trial to trial.
Instead of using a combination, as in the case of the binomial probability, the number of
ways the outcomes can occur is done using distinguishable permutations.
An example here will be much more useful than a formula.
The probability that a person will pass a College Algebra class is 0.55, the probability that
a person will withdraw before the class is completed is 0.40, and the probability that a
person will fail the class is 0.05. Find the probability that in a class of 30 students, exactly
16 pass, 12 withdraw, and 2 fail.
Outcome
x
p Ioutcome L
Pass
16
Withdraw 12
Fail
2
Total
30
0.55
0.40
0.05
1.00
The probability is found using this formula:
P=
12.
30!
H16!L H12!L H2!L
* 0.5516 * 0.4012 * 0.052 (Count this ;) )
The odds for a serial producted single-use item having one error is 7%, two
errors is 3% and no errors is 90%. What is the probability in a set of hundred
items
a) there are exactly 90 products without errors, 7 products with 1 error and 3
products with 2 errors,
b) at least one product having at least one error?
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Hypergeometric Probabilities
Open ê Close
Hypergeometric Probabilities
Hypergeometric experiments occur when the trials are not independent of each other and
occur due to sampling without replacement -- as in a five card poker hand (see the Solved
Problem 5.5 later).
Hypergeometric probabilities involve the multiplication of two combinations together and
then division by the total number of combinations.
Example 1.
How many ways can 3 men and 4 women be selected from a group of 7 men and 10
women?
7 10
3
4
The answer is =
= 7350/19448 º 0.3779
17
7
Note that the sum of the numbers in the numerator are the numbers used in the
combination in the denominator.
This can be extended to more than two groups and called an extended hypergeometric
problem.
Five cards are dealt from the deck.
What is the probability that you get
a)
a pair
b)
two pairs
c)
three of a kind
d)
straight
e)
flush
f)
full house
g)
four of a kind
h)
straight flush?
Solution:
13
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4
12
4
4
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a)P(”a pair”) =
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
13
1
4
2
12
3
4
1
4
1
4
1
11
1
4
1
52
5
n
= Binomial[n,k]
k
b) P(”two pairs”) =
13
2
4
2
c) P(”three of a kind”) =
4
2
52
5
13 4
1
3
12
2
4
1
4
1
52
5
d) Note. also includes a straight flush P(”straight”) =
10 * 45
52
5
e) Note. also includes a straight flush P(”flush”) =
4
1
13
5
52
5
f ) P(”full house”) =
13
1
g) P(”four of a kind”) =
h) P(”straight flush”) =
4
3
12
1
52
5
13 4
1
4
4
2
12
1
4
1
52
5
4 * 10
52
5
a)
HBinomial@13, 1D ∗ Binomial@4, 2D ∗ Binomial@12, 3D ∗ Binomial@4, 1D ∗
1
Input @
Binomial@4, 1D ∗ Binomial@4, 1DL ∗
Binomial@52, 5D
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Input @
352
833
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
êê N
0.422569
b)
HBinomial@13, 2D ∗ Binomial@4, 2D ∗ Binomial@4, 2D ∗
1
Input @
Binomial@11, 1D ∗ Binomial@4, 1DL ∗
Binomial@52, 5D
Input @
198
4165
êê N
0.047539
c)
HBinomial@13, 1D ∗ Binomial@4, 3D ∗ Binomial@12, 2D ∗
1
Input @
Binomial@4, 1D ∗ Binomial@4, 1DL ∗
Binomial@52, 5D
Input @
88
4165
êê N
0.0211285
d)
Input @
Input @
10 ∗ 45
Binomial@52, 5D
128
32 487
êê N
0.00394004
e)
Input @
Input @
HBinomial@4, 1D ∗ Binomial@13, 5DL ∗
33
16 660
1
Binomial@52, 5D
êê N
0.00198079
f)
HBinomial@13, 1D ∗ Binomial@4, 3D ∗
Input @
Input @
Binomial@12, 1D ∗ Binomial@4, 2DL ∗
6
4165
1
Binomial@52, 5D
êê N
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0.00144058
g)
HBinomial@13, 1D ∗ Binomial@4, 4D ∗
Input @
Binomial@12, 1D ∗ Binomial@4, 1DL ∗
1
Input @
4165
1
Binomial@52, 5D
êê N
0.000240096
h)
10 ∗ 4
Input @
Binomial@52, 5D
1
Input @
êê N
64 974
0.0000153908
Discrete Distributions
Open ê Close
Random variable's x value is ruled by chance.
-A random variable is discrete or in other words discontinious, if it can get only exact
values.
-A random variable is continuous, if it can get any values.
Example 1.
Define a random variable x's="the sum of two die rolls" probability distribution and
cumulative distribution function. Calculate the expected value and the standard deviation
of the distribution. Present the distribution graphically.
x Calc.
Prob.
Cumul.
2
Å16
Å16
1
36
1
36
3
Å26 * Å16
2
36
3
36
4
Å36 * Å16
3
36
6
36
5
Å46 * Å16
Å46
10
36
6
Å56 * Å16
5
36
15
36
7
Å66 * Å16
6
36
21
36
*
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8
Å56 * Å16
5
36
26
36
9
Å46 * Å16
4
36
30
36
10
Å36 * Å16
3
36
33
36
11
Å26 * Å16
2
36
35
36
12
Å16 * Å16
Å16
36
36
S
-
1
1
The expected value m = 2 *
1
+ 3*
36
n
-
S Jxi - xN
i-1
s=
n-1
=
2
36
+4*
3
...= 7
36
H2 - 7L2 + 2 * H3 - 7L2 + 3 * H4 - 7L2 + ... + H12 - 7L2
º 4.971
35
Five cards are drawn from a deck of cards. Form a distribution for a random
variable x, which tells the number of the cards from the most frequently drawn
suit.
14.
p
x
-
2 B
3
4
1
13
2
4
B
1
3 13
3
1
13 3
3
2
4
5
13 13
1
1
13 13
1
1
4 13
B
1
4
4
1
⁄
4 13 3
1
2
2
4 13 3
+
1
3
1
3 13
52
Fì
F
1
1
5
13
52
ì
5
5
+
~p
13 13
52
Fì
2
1
5
13
52
Fì
2
5
~ 62.89 %
~ 32.62 %
~ 4.29 %
~ 0.198 %
1
Binomial distribution
Open ê Close
x~Bin(n,p)
Expected value m = np
Standard deviation s =
npq
Example 1.
A die is rolled five times. Form a distribution for the random variable x that tells the
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frequency of sixes. Calculate the expected value and the standard deviation. Display the
distribution graphically.
Input @
n = 5;
1
p =
;
6
MDSBinomialDistributionPlot@8n, p<,
MD2σRegion → TrueD
Binomial Distribution with n = 5, p = 0.166667
Probability
0.4
0.3
0.2
0.1
0
1
2
3
4
5
X
Binomial Distribution
µ
σ
P Hµ − 2 σ ≤ X ≤ µ + 2 σL
0.833333 0.833333
0.964506
5
Answer: Both the expected value and the standard deviation are Å
6
Example 2.
Approximately 8% of Finnish men suffer from color blindness. What is the probability
that out of schools 400 boys...
a) 2 or less suffer from color blindness
b) atleast two suffer from color blindness
n = 400 ;
p = 0.08 ;
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x = 2;
MDSBinomialDistributionPnp@ X ≤ x , 8n, p<D
H∗ PH X ≤ value L = ? ∗L
PHX § 2L RowBox@8<, , SuperscriptBox@10, RowBox@8-, 9<DD<D
Probability
0.07
0.06
0.05
0.04
0.03
0.02
0.01
99
199
299
399
Binomial Distribution
n
p
µ
σ
PH X ≤ 2L
400 0.08 32. 5.42586 2.0929443760062278182 × 10−12
Input @
n = 400 ;
p = 0.08 ;
x = 3;
MDSBinomialDistributionPnp@ x ≤ X , 8n, p<D
H∗ PH value ≤ X L = ? ∗L
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PH3 § XL > 0.9999
Probability
0.07
0.06
0.05
0.04
0.03
0.02
0.01
99
199
299
399
Binomial Distribution
n
p
µ
σ
PH 3 ≤ XL
400 0.08 32. 5.42586 0.99999999999790705562
Answer: The probability of having 2 or less suffering from color blindness is around 0.
Atleast two suffer from color blindness is around 1.
Poisson distribution
Open ê Close
Named after the French mathematician Simeon Poisson, Poisson probabilities are useful
when there are a large number of independent trials with a small probability of success on
a single trial and the variables occur over a period of time. It can also be used when a
density of items is distributed over a given area or volume.
In a repeated trial the Poissson distribution is used instead of the binomial distribution,if
-the exact values of the parameters n and p are not known
-the number of trials is great and the probability of the examined event A is small
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x ~ Poisson(a), a is the expected value
P(x = k) =
ak -a
e
k!
If there are 500 customers per eight-hour day in a check-out lane, what is the probability
that there will be exactly 3 in line during any five-minute period?
The expected value during any one five minute period would be 500 / 96 = 5.2083333. The
96 is because there are 96 five-minute periods in eight hours. So, you expect about 5.2
customers in 5 minutes and want to know the probability of getting exactly 3.
P(X = 3) =
K 500 O
96
3
-500
e 96
3!
λ = 3;
x = 3;
Input @
N@PDF@PoissonDistribution@λD, xDD
H∗ PHX = xL=? ∗L
0.224042
Example 1.
The average European Football Championship final tournament game witnessed 2,5
goals. What was the probability that in a randomly picked game...
a)three goals were made?
b)atleast 2 goals were made?
a)P(x = 3) =
2.53
‰ -2.5
3!
λ = 2.5;
x = 3;
Input @
N@PDF@PoissonDistribution@λD, xDD
H∗ PHX = xL=? ∗L
0.213763
b)P("atleast two goals")=1- P("one or no goals")=1-[
2.50
0!
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‰
-2.5
+
2.51
‰ -2.5 ]º0.71
1!
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λ = 2.5 ;
x = 2;
MDSPoissonDistributionP@ x ≤ X , 8λ<D
H∗ PH value ≤ X L = ? ∗L
PH2 § XL º0.71
Probability
0.25
0.20
0.15
0.10
0.05
1
3
5
7
9
11
13
X
Poisson Distribution
λ
µ
σ
PH 2 ≤ XL
2.5 2.5 1.58114 0.71270250481635421691
Answer: The probability of 3 goals was 0.21
The probability of atleast two goals was 0.71
Example 2
Merja gets on average 20 text messages a day (12h). What is the probability of Merja
getting atleast three text messages during the next hour.
P("3 or more text messages") = 1 - P("two or less goals") =
1-[
J 20 N
12
0
20 1
20 J N
12
‰ 12 +
0!
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1!
20
‰ 12
+
J 20 N
12
2
-
20
‰ 12 ]
2!
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λ =
Input @
20
1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb
;
12
x = 3;
MDSPoissonDistributionP@ x ≤ X , 8λ<D
H∗ PH value ≤ X L = ? ∗L
PH3 § XL º0.23
Probability
0.30
0.25
0.20
0.15
0.10
0.05
1
3
5
7
9
X
Poisson Distribution
λ
µ
σ
PH 3 ≤ XL
1.66667 1.66667 1.29099 0.23400449960322143291
Answer: The probability of Merja getting atleast 3 text messages during the next hour is
23.4%
15.
The number of calls to a call center follows the Poisson distribution. Calculate
the probability that the call center receives atleast 5 calls a minute, when the
center's call frequency a is 3.
P("atleast 5 calls a minute")=1-P("4 or less calls a minute") =
1-[
30 -3 31 -3 32 -3 33 -3 34 -3
‰ +
‰ + ‰ + ‰ + ‰ ]
0!
1!
2!
3!
4!
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λ =3;
x = 5;
Input @
MDSPoissonDistributionP@ x ≤ X , 8λ<D
H∗ PH value ≤ X L = ? ∗L
PH5 § XL º0.18
Probability
0.20
0.15
0.10
0.05
1
3
5
7
9
11
13
X
15
Poisson Distribution
λ µ
σ
PH 5 ≤ XL
3. 3. 1.73205 0.18473675547622793371
Answer: The probability of the call center getting atleast 5 calls a minute is 18%
A new hamburger bar is designed to serve four customers a minute. During the
lunch hour of the first day the bar received three customers a minute on
average. Approximate the probability of demand exeeding the service capasity
during the lunch hour of the next day.
16.
a=3
The demand exeeds the service capasity, if the bar gets more than 4 customers a minute.
P("more than 4 customers") = 1 - P("4 or less customers") =
1-[
30
0!
‰ -3 +
31 -3 32 -3 33 -3 34 -3
‰ ]
‰ + ‰ + ‰ +
1!
2!
3!
4!
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λ =3;
x = 5;
MDSPoissonDistributionP@ x ≤ X , 8λ<D
H∗ PH value ≤ X L = ? ∗L
PH5 § XL º0.18
Probability
0.20
0.15
0.10
0.05
1
3
5
7
9
11
13
15
X
Poisson Distribution
λ µ
σ
PH 5 ≤ XL
3. 3. 1.73205 0.18473675547622793371
Answer: The probability of demand exeeding service capacity is 18%
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Continuous Distributions
Open ê Close
Example 1.
Busses arrive to a bus stop in 15 min intervals. A person who doesn't know the bus
schedule arrives to a stop at a random time. What is the probability that...
a)... the person has to wait for a maximum of five minutes
b)...the person hast to wait for a mimimum of eigth minutes
c)...3-6 min
d)exactly 10 minutes
e)What is the expected value of the wait?
a)
b)
5
1
= Å
15 3
7
15
3
1
= Å
15 5
d) 0
e) 7.5 min
c)
17.
1
By what value of the constant b, f(x) = 1 - Å x, when 0 § x § b, otherwise 0, is a
8
probability density function?
Then calculate P(x ¥ 1)
A=1
Input @
MDPlotB:1 −
Site: www.deltasoft.at
1
8
x,>, 8x , 0 , 8 < F
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y
1
0.8
0.6
0.4
0.2
x
2
1 * K1 - Å1 bO
8
4
6
8
*b = 1
2
-
1 2 1
b + Å b-1=0
16
2
Æ Pure Solve ;
Input @
Clear@xD;
MDRealOnlyBSolveB: −
1
16
b2 +
1
2
b − 1 == 0>, 8 b < FF
88b → 4<, 8b → 4<<
Answer: b is 4.
The side length of a square shaped room is 4m. The random variable x is the
distanse of a glass bead dropped on the room surface from the closest wall.
18.
a) Define P(x § 0.5)
b) Define the random variable's cumulative distribution function and plot it
c) Define the random variable's probability density function and plot it
a)
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The area of the gray part of the room is 42 - H4 - 2 xL2 = 16 - H4 - 2 xL2 .
The probability that x § 0.5 is
16 - H4 - 2 * 0.5L2
16
Input @
16 − H4 − 2 ∗ 0.5L2
16
0.4375
b) The cumulative distribution function is
I16 - H4 - 2 xL2 M
when 0 § x § 2, 1 when x > 2,
16
and 0 when x < 0.
Input @
Clear@xD;
Input @
F@x_D = I16 − H4 − 2 xL2 M ë 16
1
16
Input @
I16 − H4 − 2 xL2 M
MDPlot@8F@xD<, 8x , 0 , 2 < D
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y
1
0.8
0.6
0.4
0.2
x
0.5
1
1.5
2
Graphics c) The density function is the derivative of the cumulative function.
Input @
F '@xD
1
4
H4 − 2 xL
when 0 § x § 2, and 0 elsewhere.
Input @
MDPlot@8F '@xD<, 8x , 0, 2 < D
y
1
0.8
0.6
0.4
0.2
x
0.5
1
1.5
2
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Normal distribution
Open ê Close
x ~ N(m,s)
The expected value is m and the standard deviation is s.
Z=
x- m
s
Example 1
Intelligence quotient is measured with a test, which average is 100 and standard
deviation is 15.
What is the probability that a randomly picked person has an IQ of under 97?
The IQ limit 97 correlates to norm value
97 - 100
= -0.2
15
P(x § -0.2) = F(-0.2) = 1 - F(0.2) = 0.42074
Æ ;
µ = 100 ; σ = 15 ; x = 97 ;
Input @ probability = MDSφ@x, 8µ, σ<D;
MDSNormalDistributionPµσ@ X ≤ x , 8µ, σ<D
H∗ PH X ≤ value L = ? ∗L
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PHX § 97L = 0.42074
40
60
80
120
140
160
X
Normal Distribution
z H−∞L z H97.L PH X ≤ 97.L
−∞
−0.2
0.42074
Dev Hz σL
100. − 3.
Answer: The probability of a randomly picked person scoring under 97 in this test is 42%.
19.
The weight of boys at the age of 18 follows the Normal distribution, so that the
average weight is 66 kg and the standard deviation is 8kg. What percentage of
18 year old boys weigh from 60 to 70 kg?
P("between 60kg and 70kg") = P("under 70kg") - P("under 60kg").
µ = 66 ; σ = 8 ; x1 = 60 ; x2 = 70 ;
probability = MDSφ@x2, 8µ, σ<D − MDSφ@x1, 8µ, σ<D;
Input @
MDSNormalDistributionPµσ@ x1 ≤ X ≤ x2, 8µ, σ<D
H∗ PH x1 ≤ X ≤ x2 L = ? ∗L
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PH60 § X § 70L = 0.464835
30
40
50
60
70
80
90
100
X
Normal Distribution
z H60.L z H70.L PH 60 ≤ X ≤ 70L
−0.75
0.5
0.464835
Answer: 46% of boys weigh between 60kg and 70kg.
Let's presume that the mass per area in paper is normally distributed. A client
orders paper which mass per area ratio is 80 g í m2 . How great can the mass
20.
z=
per area ratio's deviation be for the probability of getting paper with less than 75
g í m2 mass per area is less than 5%?
x- m
s
,
s=
x- m
z
probability = 0.05 ;
Clear@zD; z = z ê. FindRoot@MDSφ@zD == probability, 8z, 0<D;
Input @
MDSNormalDistributionPµσ@ Z ≤ z, 80, 1<D
H∗ PH Z ≤ ? L = probab ∗L
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PHZ § -1.64485 L = 0.05
-4
-2
2
4
Z
Normal Distribution
z H−1.64485L PH Z ≤ −1.64485L
−1.64485
0.05
z H−∞L
−∞
Dev Hz σL
0. −1.64485
Input @
σ=
75 − 80
−1.64485
3.03979
Answer: The standard deviation can be 3.04 g/m2 at maximum.
21.
One can get in to a society of top intelligent people, if one's intelligence is
greater than that of 98% of the population. What IQ does one have to have to
get into this society? IQ~(100,24)
The z of 98%
probability = 0.98 ;
Clear@zD; z = z ê. FindRoot@MDSφ@zD == probability, 8z, 0<D;
Input @
MDSNormalDistributionPµσ@ Z ≤ z, 80, 1<D
H∗ PH Z ≤ ? L = probab ∗L
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P HZ § 2.05375L = 0.98
-4
-2
Z
2
4
Normal Distribution
z H−∞L
z H2.05375L P H Z ≤ 2.05375L
−∞
2.05375
0.98
Dev Hz σL
0. +2.05375
Input @
µ = 100
100
Input @
σ = 24
24
Input @
x = z∗σ + µ
149.29
Answer: For one to get in one has to have an IQ over 149.
If xi (i = 1,2,3,... ,n) are normally distributed random variable's, which averages
are m i and standard deviations are s i , so likewise x1 +x2 +x3 +...+ xn are N(m,s 2 )
where m= m 1 + m 2 + m 3 +... m n and s 2 =s 2 1 +s 2 2 +s 2 3 +...+s 2 n
22.
Solve the following problem: There are 23 sweets in the bag. The mass of the
sweeties is normaly distributed, so that the average is 2,1g and the standard
deviation is 0,25g. What is the probability that the sweets in the bag weigh in
together at over 50g?
Æ ;
µ = 23 ∗ 2.1 ; σ = 23 ∗ 0.252 ; X = 50 ;
Input
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23 2.1 ;
z =
X−µ
σ
23 0.25 ; X
50 ;
êê N
1.18261
Æ ;
z = 1.1826086956521709` ;
Input @ probability = 1 − MDSφ@zD;
MDSNormalDistributionPµσ@ z ≤ Z, 80, 1<D
H∗ PH z ≤ Z L = ? ∗L
P H1.18261 § ZL = 0.118482
-4
-2
Z
2
4
Normal Distribution
z H1.18261L z H+∞L P H 1.18261 ≤ ZL
1.18261
+∞
0.118482
Dev Hz σL
0. +1.18261
Answer: The probability of the bags weight topping 50 g is 11.8%.
The mean velocity of Finnish traffic flow is 81 km/h with in 80 km/h area. 13.4%
of drivers speed 10km/h or more over the allowed 80 km/h limit.
23.
a) Calculate the standard deviation of velocities, when the velocities are
presumed to be normaly distributed.
b) How great portion of the drivers don't survive with a traffic fine, but get a
penalty demand in other words speed over 21 km/h?
a) Now P(x § 90) = 1 - 0.134 = 0.866
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probability = 0.866 ;
Clear@zD; z = z ê. FindRoot@MDSφ@zD == probability, 8z, 0<D;
MDSNormalDistributionPµσ@ Z ≤ z, 80, 1<D
H∗ PH Z ≤ ? L = probab ∗L
PHZ § 1.10768 L = 0.866
-4
-2
2
4
Z
Normal Distribution
z H−∞L
z H1.10768L PH Z ≤ 1.10768L
−∞
1.10768
0.866
Dev Hz σL
0. +1.10768
Input @
σ=
90 − 81
1.10768
8.12509
b) We solve the standardized value of speeding 21km/h. (Driving at 101 km/h)
z=
x- m
Input @
s
z=
101 − 81
8.125
2.46154
Æ ;
z = 2.4615384615384617` ;
Input @ probability = 1 − MDSφ@zD;
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probability
1
MDS @zD;
MDSNormalDistributionPµσ@ z ≤ Z, 80, 1<D
H∗ PH z ≤ Z L = ? ∗L
P H2.46154 § ZL = 0.00691713
-4
Z
-2
2
4
Normal Distribution
z H2.46154L z H+∞L P H 2.46154 ≤ ZL
2.46154
+∞
0.00691713
Dev Hz σL
0. +2.46154
Answer: The standard deviation is 8.1 km/h
0.69% of the drivers speed over 21 km/h.
Replacing the binomial distribution
with the normal distribution
Open ê Close
When number of repeats n is great the binomial distribution Bin(n,p) can be replaced
with the normal distribution N(m,s), where m = np and
s=
npq
Example 1.
The probability of a customer returning a product he/she has ordered from mailorder is
11%. What is the probability, that out of mailorder company's 2500 customers 300 or
more return the products they've ordered?
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n = 2500;
Input @ p = 0.11;
:µ = n p , σ =
n p H1 − pL > êê N
8275., 15.6445<
Let's standardize 300,
Æ ;
µ = 275; σ = 15.644487847162015 ; X = 300 ;
Input @
z =
X−µ
σ
êê N
1.59801
Æ ;
z = 1.59801 ;
Input @ probability = 1 − MDSφ@zD;
MDSNormalDistributionPµσ@ z ≤ Z, 80, 1<D
H∗ PH z ≤ Z L = ? ∗L
P H1.59801 § ZL = 0.0550204
-4
-2
Z
2
4
Normal Distribution
z H1.59801L z H+∞L P H 1.59801 ≤ ZL
1.59801
+∞
0.0550204
Dev Hz σL
0. +1.59801
Answer: The probability of 300 or more customers returning their product is 5.5%
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About 5% of people are left handed. What is the probability, that there are
atleast 30 "lefties" in Lyseonpuiston lukio (Lyska) ? (circa 800 students)
n = 800;
Input @ p = 0.05;
:µ = n p , σ =
n p H1 − pL > êê N
840., 6.16441<
Æ ;
µ = 40 ; σ = 6.164414002968976` ; X = 30 ;
Input @
z =
X−µ
σ
êê N
−1.62221
Æ ;
z = −1.6222142113076254` ;
Input @ probability = 1 − MDSφ@zD;
MDSNormalDistributionPµσ@ z ≤ Z, 80, 1<D
H∗ PH z ≤ Z L = ? ∗L
P H-1.62221 § ZL = 0.947621
-4
-2
Z
2
4
Normal Distribution
z H−1.62221L z H+∞L P H −1.62221 ≤ ZL
−1.62221
+∞
0.947621
Dev Hz σL
0. −1.62221
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Answer: The probability of 'Lyska' having atleast 30 "lefties" is 94.8%.
25.
A presidential candidate is supported by 48,4% of the voters. Calculate the
probability, that the majority out of randomly picked group of 1000 people
participating in an election poll are going to support our covered candite?
n = 1000;
Input @ p = 0.484;
:µ = n p , σ =
n p H1 − pL > êê N
8484., 15.8033<
Æ ;
µ = 484.` ; σ = 15.803290796539814` ; X = 500;
Input @
z =
X−µ
σ
êê N
1.01245
Æ ;
z = 1.0124473570721901` ;
Input @ probability = 1 − MDSφ@zD;
MDSNormalDistributionPµσ@ z ≤ Z, 80, 1<D
H∗ PH z ≤ Z L = ? ∗L
P H1.01245 § ZL = 0.155662
-4
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-2
Z
2
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Normal Distribution
z H1.01245L z H+∞L P H 1.01245 ≤ ZL
1.01245
+∞
0.155662
Dev Hz σL
0. +1.01245
Answer: The probability of the candidate getting the majority in the poll is 15.6%.
Sample selection
Open ê Close
An avarege calculated from a sample is an estimate for the real distributions average
i.e expected value. When we pick multiple same size samples the same variable in
mind and calculate the average of each sample, results are very often normaly
distributed. Even when the variable itself is not normaly distributed. In this kind of
situation we use the average calculated from the sample as the expected value x and
s
as standard deviation we use the average's mean error
, where n is the size of the
n
sample. X and s are indexes of the random variable's distribution or if they are not
known like the case often is, they are indexes calculated from the sample.
Example 1.
The average weight of salmon fished out from the river Teno was 8,5 kg and the
salmon's weight's deviation was 3 kg.
What is the probability that...
a) the next fish weighs over 11kg?
b) the average weight of the next 20 fishes is over 10 kg?
a)
Æ ;
µ = 8.5 ; σ = 3 ; x = 11 ;
Input @ probability = 1 − MDSφ@x, 8µ, σ<D;
MDSNormalDistributionPµσ@ x ≤ X, 8µ, σ<D
H∗ PH value ≤ X L = ? ∗L
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P H11 § XL = 0.202328
-5
X
0
5
10
15
20
Normal Distribution
z H11.L z H+∞L P H 11. ≤ XL
0.833333
+∞
0.202328
Dev Hz σL
8.5 + 2.5
b)
Æ ;
µ = 8.5 ; σ =
3
; x = 10 ;
20
Input @
probability = 1 − MDSφ@x, 8µ, σ<D;
MDSNormalDistributionPµσ@ x ≤ X, 8µ, σ<D
H∗ PH value ≤ X L = ? ∗L
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P H10 § XL = 0.0126737
X
6
7
8
9
10
11
Normal Distribution
z H10.L z H+∞L P H 10. ≤ XL
2.23607
+∞
0.0126737
Dev Hz σL
8.5 + 1.5
Answer: The probability of the next fish weighing over 11kg is 20.2%.
The probability of the average of the next 20 being over 10 kg is 1.3%.
26.
The average service time in a hamburger bar is 8 minutes and the standard
deviation is 3 minutes. The bar has ten customers. Calculate the probability that
the average service time for these customers exeeds 10 minutes.
Æ ;
µ = 8; σ =
3
; x = 10;
10
Input @
probability = 1 − MDSφ@x, 8µ, σ<D;
MDSNormalDistributionPµσ@ x ≤ X, 8µ, σ<D
H∗ PH value ≤ X L = ? ∗L
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P H10 § XL = 0.0175075
X
4
6
10
12
Normal Distribution
z H10.L z H+∞L P H 10. ≤ XL
2.10819
+∞
0.0175075
Dev Hz σL
8. + 2.
Answer: the probability is 1.8%
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Summary
Open ê Close
Print
Mathematica project summary, what have we learnt:
During this project we have learnt how to use Mathematica program and to cooperate with
students from other countries from all around Europe. One of most notable gains we have
also gotten from the project was learning basic mathematical English. This was due to
numerous social interactions with students and teachers from different cultural
backgrounds. Interaction between different cultures taught us also valuable lessons from
other than mathematical fields.
Algebra group thought that the most valuable usage of the Mathematica and M@th
Desktop is in graph drawing, it enables to imagine problems in mathematics better.
Functional group emphasizes that the most remarkable aspect of Mathematica and M@th
Desktop is that it concretizes mathematics. For example by drawing the functional exercise
it was easier to perceive the problem. The program was very flexible and it made it
possible to solve difficult exercises in short time.
Students in trigonometry group have learnt many things about how to use Mathematica
and M@th Desktop program. The group discovered that the program is a useful tool to
solve many different types of trigonometrical exercises.
Probability group focused on probability in which they improved a lot during the project.
The group learned of the several useful and efficient features of Mathematica and M@th
Desktop for probability problems.
E-learning was one of the main focuses of the project and it was easy and still effective to
solve difficult exercises with international teamwork. All groups think that the best part of
the project was the necessity to communicate with students from foreign countries,
especially including language practice, which still continues. Most of us have kept contact
with each other also outside the project and we have made friends with them. Mathematica
project has created a new aspect to our mathematical study techniques.
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Our Team
Open ê Close
Print
Gymnazium Teplice, Teplice, Czech Republic
Jakub Krmela
Zuzana Jaklinova
David Kubon
Jan Kubalik
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Jan Moravec
Eva Trubenekrova
Jan Hodek
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Lyseopuistonlukio, Rovaniemi, Finland
Tiina Alaruikka
Petri Huttu
Iida Impiö
Annemari Kiviniemi
Essi-Riina Korva
Anna Maijala
Merja Mettänen
Eero Pihkala
Antti Pyykkönen
Paulus Sieppi
Tommi Tolonen
Site: www.deltasoft.at
Miika Haataja
Leena Hytönen
Henri Izadi
Sami Kojola
Laura Lehtiniemi
Johan Matinmikko
Kalle Niemi
Tanja Pohjola
Laura Sandqvist
Noora Siitonen
Riina Typpö
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The ZST Technical School, Mikolow, Poland
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Gymnazium A. Bernolaka, Namestovo, Slovakia
New Chapter
Cut Last Chapter
Open ê Close
Print
List of our sources:
We have taken some parts of the Linear function, Quadratic Functions and Trigonometric
Functions notebook of the e-learning software M@th Desktop 5.0 and edited it.
M@th Desktop 5.0 is based on Mathematica.
PCMLogo: PCM Homepage
Article: lllll
Formula: Booktitle, Author, Publ. Comp, Year, jméno knihy dopsat !!!
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