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LESSON
Name
Date
6.1 Study Guide
For use with pages 269–274
GOAL
Find ratios and unit rates.
VOCABULARY
A ratio uses division to compare two quantities.
Two ratios are called equivalent ratios when they have the same value.
Lesson 6.1
EXAMPLE
1 Writing Ratios
In tennis class, you have won 15 out of 24 matches. You have lost 9 matches.
Write the ratio in three ways.
a. The number of wins to the number of losses
b. The number of wins to the total number of matches
Solution
Number of wins
Number of losses
15
9
5
3
Number of wins
Total number of matches
a. ᎏᎏ ⫽ ᎏᎏ ⫽ ᎏᎏ
15
24
5
8
b. ᎏᎏᎏ ⫽ ᎏᎏ ⫽ ᎏᎏ
Three ways to write the ratio
Three ways to write the ratio are
5
are ᎏᎏ, 5 to 3, and 5 : 3.
3
5
ᎏᎏ, 5 to 8, and 5 : 8.
8
Exercise for Example 1
1. Using the tennis information above, compare the total number of matches
to the number of matches lost using a ratio. Write the ratio in three ways.
EXAMPLE
2 Comparing and Ordering Ratios
Order the ratios 5 : 11, 13 : 23, and 3 : 7 from least to greatest.
Solution
Write each ratio as a fraction. Then use a calculator to write each fraction as a
decimal. Round to the nearest hundredth and compare the decimals.
5
ᎏᎏ ≈ 0.45
11
13
ᎏᎏ ≈ 0.57
23
3
ᎏᎏ ≈ 0.43
7
Answer: The ratios, from least to greatest, are 3 : 7, 5 : 11, and 13 : 23.
Exercises for Example 2
Order the ratios from least to greatest.
8
11
2. 5 to 12, ᎏᎏ, 3 : 5
Copyright © McDougal Littell/Houghton Mifflin Company
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1
3
3. ᎏᎏ, 3 : 10, 21 to 65
7
12
4. ᎏᎏ, 5 to 11, 18 : 37
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Name
Date
6.1 Study Guide
Continued
EXAMPLE
For use with pages 269–274
3 Finding a Unit Rate
You pay $3.45 for 15 ounces of cereal. What is the cost per ounce?
Solution
First, write a rate comparing the total cost of the cereal to the total amount
of cereal. Then rewrite the rate so the denominator is 1.
Lesson 6.1
$3.45
$3.45 ⫼ 15
ᎏᎏ ⫽ ᎏᎏ
15 oz
15 oz ⫼ 15
Divide numerator and denominator by 15.
$.23
1 oz
⫽ ᎏᎏ
Simplify.
Answer: The cereal costs $.23 per ounce.
EXAMPLE
4 Using Equivalent Rates
A proposed hypersonic plane will be able to travel at a rate of 5040 miles per
hour. How many feet will the plane be able to travel in 0.5 second?
Solution
(1) Express the plane’s rate in miles per second.
1m
冫
in
1 h冫
5040 mi
5040 mi
1.4 mi
ᎏᎏ ⫽ ᎏᎏ p ᎏᎏ p ᎏᎏ ⫽ ᎏᎏ
6
0
s
e
c
1h
sec
1 h冫
60 冫
min
Multiply by conversion factors, divide
out common units, and simplify.
(2) Find the distance (in feet) that the plane travels in 0.5 second.
Distance ⫽ Rate p Time
Write formula for distance.
Substitute values. Divide out
common unit.
Multiply.
1.4 mi
sec
冫
冫
⫽ ᎏᎏ p 0.5 sec
⫽ 0.7 mi
⫽ 0.7 冫
mi p ᎏᎏ
Multiply by conversion factor.
Divide out common unit.
⫽ 3696 ft
Simplify.
5280 ft
冫
mi
Answer: The plane will be able to travel 3696 feet in 0.5 second.
Exercises for Examples 3 and 4
Find the unit rate.
60 songs
5 hours
$.50
10 min
5. ᎏᎏ
$1315
20 people
6. ᎏᎏ
7. ᎏᎏ
$17.60
5 yd
8. ᎏᎏ
Write the equivalent rate.
140 mi
1h
? ft
1 min
9. ᎏᎏ ⫽ ᎏᎏ
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Pre-Algebra
Chapter 6 Resource Book
90 km
1h
? cm
1 sec
10. ᎏᎏ ⫽ ᎏᎏ
$3600
1 day
? dollars
1 min
11. ᎏᎏ ⫽ ᎏᎏ
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LESSON
Name
Date
6.2 Study Guide
For use with pages 275–279
GOAL
Write and solve proportions.
VOCABULARY
A proportion is an equation that states that two ratios are equivalent.
EXAMPLE
1 Solving a Proportion Using Equivalent Ratios
x
7
Solve the proportion ᎏ1ᎏ0 ⫽ ᎏᎏ.
50
(1) Compare denominators.
7
ᎏᎏ
10
(2) Find x.
x
ᎏᎏ
50
⫻5
7
ᎏᎏ
10
⫻5
x
ᎏᎏ
50
Answer: Because 7 ⫻ 5 ⫽ 35, x ⫽ 35.
Exercises for Example 1
Lesson 6.2
Use equivalent ratios to solve the proportion.
x
24
15
8
8
11
1. ᎏᎏ ⫽ ᎏᎏ
EXAMPLE
x
44
2. ᎏᎏ ⫽ ᎏᎏ
x
34
3
17
4. ᎏᎏ ⫽ ᎏᎏ
x
16
8. ᎏᎏ ⫽ ᎏᎏ
3. ᎏᎏ ⫽ ᎏᎏ
x
39
12
13
12
15
x
35
2 Solving a Proportion Using Algebra
x
15
27
45
Solve the proportion ᎏᎏ ⫽ ᎏᎏ. Check your answer.
x
27
ᎏᎏ ⫽ ᎏᎏ
15
45
x
15
Write original proportion.
27
45
15 p ᎏᎏ ⫽ 15 p ᎏᎏ
405
45
Multiply each side by 15.
x ⫽ ᎏᎏ
Simplify.
x⫽9
Divide.
x
15
27
45
✓ Check: ᎏᎏ ⫽ ᎏᎏ
9
27
ᎏᎏ ⫽ ᎏᎏ
15
45
3
3
ᎏᎏ ⫽ ᎏᎏ ✓
5
5
Write original proportion.
Substitute 9 for x.
Simplify. Solution checks.
Exercises for Example 2
Use algebra to solve the proportion.
x
4
x
9
5. ᎏᎏ ⫽ ᎏᎏ
6. ᎏᎏ ⫽ ᎏᎏ
15
10
12
18
Copyright © McDougal Littell/Houghton Mifflin Company
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15
20
7. ᎏᎏ ⫽ ᎏᎏ
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LESSON
Name
Date
6.2 Study Guide
Continued
EXAMPLE
For use with pages 275 –279
3 Writing and Solving a Proportion
For every $25 that you earn, you save $15. How much of the $70 that you
earned at your after-school job this week will you save?
Solution
First, write a proportion involving two ratios that compare the amount you save
to the amount you earn.
Amount saved
15
x
ᎏᎏ ⫽ ᎏᎏ
Amount earned
25
70
Then, solve the proportion.
15
25
x
70
70 p ᎏᎏ ⫽ 70 p ᎏᎏ
1050
ᎏᎏ ⫽ x
25
42 ⫽ x
Multiply each side by 70.
Simplify.
Divide.
Lesson 6.2
Answer: You will save $42 out of the $70 you earned this week.
Exercise for Example 3
9. At a bookstore, 3 novels cost $17.97. How many novels can you buy for
$29.95?
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LESSON
Name
Date
6.3 Study Guide
For use with pages 280–284
GOAL
Solve proportions using cross products.
VOCABULARY
Every pair of ratios has two cross products. A cross product of two
ratios is the product of the numerator of one ratio and the denominator
of the other ratio. If the cross products are equal, then the ratios form
a proportion.
EXAMPLE
1 Determining if Ratios Form a Proportion
Tell whether the ratios form a proportion.
8 10
12 15
15 20
18 25
a. ᎏᎏ, ᎏᎏ
b. ᎏᎏ, ᎏᎏ
Solution
a.
8
10
ᎏᎏ ⱨ ᎏᎏ
12
15
Write proportion.
8 p 15 ⱨ 12 p 10
Form cross products.
120 ⫽ 120
Multiply.
Answer: The ratios form a proportion.
b.
15
20
ᎏᎏ ⱨ ᎏᎏ
18
25
Write proportion.
15 p 25 ⱨ 18 p 20
375 ⫽ 360
Form cross products.
Multiply.
Lesson 6.3
Answer: The ratios do not form a proportion.
Exercises for Example 1
Tell whether the ratios form a proportion.
16 12
20 16
1. ᎏᎏ, ᎏᎏ
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All rights reserved.
15 3
75 12
2. ᎏᎏ, ᎏᎏ
18 27
32 48
25 10
65 32
3. ᎏᎏ, ᎏᎏ
4. ᎏᎏ, ᎏᎏ
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LESSON
Name
Date
6.3 Study Guide
Continued
For use with pages 280 –284
CROSS PRODUCTS PROPERTY
Words
The cross products of a proportion are equal.
6
15
2
5
Numbers Given that ᎏᎏ ⫽ ᎏᎏ, you know that 2 p 15 ⫽ 5 p 6.
a
b
c
d
Algebra If ᎏᎏ ⫽ ᎏᎏ, where b ⫽ 0 and d ⫽ 0, then ad ⫽ bc.
EXAMPLE
2 Writing and Solving a Proportion
You need 1.75 cups of lemon juice to make 14 servings of lemonade. How many
cups of lemon juice are needed to make 4 servings of lemonade?
Solution
Cups of lemon juice
Servings
1.75
c
ᎏᎏ ⫽ ᎏᎏ
4
14
1.75 p 4 ⫽ 14c
Cross products property
7 ⫽ 14c
Multiply.
7
14c
ᎏᎏ ⫽ ᎏᎏ
14
14
Divide each side by 14.
0.5 ⫽ c
Simplify.
Answer: You need 0.5 cup of lemon juice to make 4 servings of lemonade.
Exercises for Example 2
Use the cross products property to solve the proportion.
a
30
2
12
b
6
8
12
6. ᎏᎏ ⫽ ᎏᎏ
0.7
c
84
15
7. ᎏᎏ ⫽ ᎏᎏ
0.64
16
2
d
8. ᎏᎏ ⫽ ᎏᎏ
Lesson 6.3
5. ᎏᎏ ⫽ ᎏᎏ
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Chapter 6 Resource Book
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LESSON
Name
Date
6.4 Study Guide
For use with pages 287–292
GOAL
Identify similar and congruent figures.
VOCABULARY
Two figures are similar figures if they have the same shape but not
necessarily the same size. The symbol S indicates that two figures are
similar. Corresponding parts of figures are sides or angles that have
the same relative position.
Two figures are congruent if they have the same shape and the same
size. If two figures are congruent, then the corresponding angles are
congruent and the corresponding sides are congruent.
EXAMPLE
1 Identifying Corresponding Parts of Similar Figures
Given TABC S TDEF, name the corresponding
angles and the corresponding sides.
B
E
D
C
F
A
Solution
Corresponding angles: aA and aD, aB and aE, aC and aF
&, &
& and EF
&
&, &
& and &
&
Corresponding sides: &*
AB and &
DE
BC
AC
DF
Exercise for Example 1
1. Given JKLMP S QRSTU, name the corresponding angles and the
corresponding sides.
K
U
J
L
R
T
P
EXAMPLE
M
S
2 Finding the Ratio of Corresponding Side Lengths
Given ABCDEF S GHJKLM, find the ratio of the
lengths of the corresponding sides of ABCDEF
to GHJKLM.
A
Lesson 6.4
E 3.6 in. D
Answer: The ratio of the lengths of the corresponding
38
Pre-Algebra
Chapter 6 Resource Book
2.7 in.
C
5.1 in.
4.2 in.
AB
6
3
ᎏᎏ ⫽ ᎏᎏ ⫽ ᎏᎏ
GH
2
1
3
1
B
F
Write a ratio comparing the lengths of a pair of
corresponding sides. Then substitute the lengths
of the sides and simplify.
sides is ᎏᎏ.
6 in.
3.3 in.
1.4 in. M 1.1 in.
G
L
2 in.
1.2 in.
K
H
1.7 in. J 0.9 in.
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LESSON
Name
Date
6.4 Study Guide
Continued
EXAMPLE
For use with pages 287–292
3 Checking for Similarity
Tell whether the triangles are similar.
L
B
Solution
Because aA c aJ, aB c aK, and aC c aL,
corresponding angles are congruent. To decide
if the triangles are similar, determine whether
the ratios of the lengths of corresponding sides
are equal.
A
6.25 m
C
K
AB
BC
AC
ᎏᎏ ⱨ ᎏᎏ ⱨ ᎏᎏ
JK
KL
JL
Write proportion.
5
5.625
6.25
ᎏᎏ ⱨ ᎏᎏ ⱨ ᎏᎏ
8
9
10
Substitute values.
0.625 ⫽ 0.625 ⫽ 0.625
Divide. The ratios are equal.
10 m
9m
5.625 m
5m
J
8m
Answer: The corresponding angles are congruent and the ratios of the lengths
of the corresponding sides are equal, so TABC S TJKL.
EXAMPLE
4 Finding Measures of Congruent Figures
Given GHJK c PQRS, find the indicated measure.
a. GH
2.2 in. J
119.7⬚
H
b. maK
Solution
Because the quadrilaterals are congruent, the corresponding
angles are congruent and the corresponding sides are congruent.
59⬚
K
5.8 in.
G
R
&
&c&
& . So, GH ⫽ PQ ⫽ 4 inches.
a. GH
PQ
116.6⬚
b. aK c aS. So, maK ⫽ maS ⫽ 64.7⬚.
3.6 in.
4 in.
64.7⬚
S
P
Exercises for Examples 2– 4
2. Tell whether the rectangles are similar. If they are similar, find the ratio of
the lengths of the corresponding sides of EFGH to ABCD.
E
5 cm
H
F
1 cm
G
A
10 cm
B
3 cm
D
C
N
S
40⬚
M
5m
120⬚
P
Copyright © McDougal Littell/Houghton Mifflin Company
All rights reserved.
Lesson 6.4
3. Given TMNP c TSTU, find TU and maN.
12.7 m
9.4 m
20⬚
U
Chapter 6
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LESSON
Name
Date
Lesson 6.5
6.5 Study Guide
For use with pages 293 –297
GOAL
EXAMPLE
Find unknown side lengths of similar figures.
1 Finding an Unknown Side Length in Similar Figures
Given JKLM S NPQR, find KL.
J
Solution
Use the ratios of the lengths of corresponding
sides to write a proportion involving the
unknown length, KL.
LM
KL
ᎏᎏ ⫽ ᎏ ᎏ
QR
PQ
x
8
ᎏᎏ ⫽ ᎏᎏ
5
2
K
x
N
P
R
2 yd
M 8 yd L
5 yd
Write proportion involving KL.
Substitute.
8 p 5 ⫽ 2x
Cross products property
40 ⫽ 2x
Multiply.
20 ⫽ x
Divide each side by 2.
& is 20 yards.
Answer: The length of &
KL
EXAMPLE
2 Using Indirect Measurement
A girl who is 150 centimeters tall
is standing next to a tree. The tree
and the girl are perpendicular to
the ground. The sun’s rays strike
the tree and the girl at the same
angle, forming two similar
triangles. The length of the girl’s
shadow is 200 centimeters, and
the length of the tree’s shadow
is 1600 centimeters. How tall is
the tree?
h
150 cm
1600 cm
200 cm
Solution
Write and solve a proportion to find the height h (in centimeters) of the tree.
Height of tree
Length of tree’s shadow
ᎏᎏ ⫽ ᎏᎏᎏ
Height of girl
Length of girl’s shadow
h
1600
ᎏᎏ ⫽ ᎏᎏ
150
200
Substitute.
200h ⫽ 150 p 1600
Cross products property
200h ⫽ 240,000
Multiply.
h ⫽ 1200
Divide each side by 200.
Answer: The height of the tree is 1200 centimeters, or 12 meters.
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LESSON
Name
Date
Continued
Lesson 6.5
6.5 Study Guide
For use with pages 293–297
Exercises for Examples 1 and 2
1. Given ABCDE S FGHJK, find CD.
33 ft
E
A
H
10 ft
J
G
D
B
x
F
K
11 ft
C
2. A 4-foot boy is standing next to a statue. The shadow cast by the boy is
3 feet and the shadow cast by the statue is 30 feet. What is the statue’s
height?
EXAMPLE
3 Using Algebra and Similar Triangles
Given TJKL S TJMN, find JN.
K
To find JN, write and solve a proportion.
KL
JL
ᎏᎏ ⫽ ᎏᎏ
MN
JN
KL
JN ⫹ NL
ᎏᎏ ⫽ ᎏᎏ
MN
JN
10
x⫹6
ᎏᎏ ⫽ ᎏᎏ
5
x
M
Write proportion.
5m
Use the fact that JL ⫽ JN ⫹ NL.
J
x
N 6m
L
Substitute.
10x ⫽ 5(x ⫹ 6)
Cross products property
10x ⫽ 5x ⫹ 30
Distributive property
5x ⫽ 30
10 m
Subtract 5x from each side.
x⫽6
Divide each side by 5.
&* is 6 meters.
Answer: The length of JN
Exercise for Example 3
3. Given TSTU S TSYZ, find ZU.
T
60 m
Y
15 m
S 20 m Z
Copyright © McDougal Littell/Houghton Mifflin Company
All rights reserved.
x
U
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LESSON
Name
Date
6.6 Study Guide
For use with pages 299–304
GOAL
Use proportions with scale drawings.
VOCABULARY
EXAMPLE
Lesson 6.6
A scale drawing is a two-dimensional drawing that is similar to the
object it represents. A scale model is a three-dimensional model that
is similar to the object it represents. The scale of a scale drawing or
scale model gives the relationship between the drawing or model’s
dimensions and the actual dimensions.
1 Using a Scale Drawing
The distance on a map from Toronto, Canada, to Ottawa, Canada, is 9 centimeters.
The scale is 1 cm : 25 mi. What is the actual distance between the two cities?
Solution
Let x represent the actual distance (in miles) between Toronto and Ottawa. The
ratio of the map distance between the two cities to the actual distance x is equal
to the scale of the map. Write and solve a proportion using this relationship.
1 cm
9 cm
Map distance
ᎏᎏ ⫽ ᎏᎏ
25 mi
x mi
Actual distance
1x ⫽ 25 p 9
Cross products property
x ⫽ 225
Multiply.
Answer: The actual distance is 225 miles.
Exercise for Example 1
1. The scale of a blueprint is 1 in. : 3 ft. The actual length of the living room
is 18 feet. Find the length of the living room on the blueprint.
EXAMPLE
2 Finding the Scale of a Drawing
A gardener is making a scale drawing of a garden. The length of the actual
garden is 9 yards. The length of the garden in the drawing is 3 inches. Find
the drawing’s scale.
Solution
Write a ratio using corresponding side lengths of the scale drawing and the
actual garden. Then simplify the ratio so that the numerator is 1.
3 in.
Length of scale drawing
ᎏᎏ
9 yd
Length of garden
3 in.
1 in.
ᎏᎏ ⫽ ᎏ ᎏ
9 yd
3 yd
Simplify.
Answer: The drawing’s scale is 1 in. : 3 yd.
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Name
Date
6.6 Study Guide
Continued
For use with pages 299 –304
Exercises for Example 2
2. The distance between two cities on a map is 12 millimeters. The actual
distance between the cities is 60 miles. Find the map’s scale.
Lesson 6.6
3. The length of a drawing of a fireplace is 10 inches. The actual length is
4 feet. Find the drawing’s scale.
EXAMPLE
3 Finding a Dimension of a Scale Model
A model of a building has a scale of 1 : 40. The building is 100 feet tall. Find
the height of the model.
Solution
Write a proportion using the scale.
1
x
Dimension of model
ᎏᎏ ⫽ ᎏᎏ
40
100
Dimension of building
100 ⫽ 40x
Cross products property
2.5 ⫽ x
Divide each side by 40.
Answer: The height of the model is 2.5 feet.
Exercises for Example 3
4. A model of a boat has a scale of 1 : 15. The model’s length is 2 feet. Find
the actual length of the boat.
5. A model of a ladybug has a scale of 1 in. : 0.3 cm. The ladybug’s actual
length is 1.2 centimeters. Find the model’s length.
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LESSON
Name
Date
6.7 Study Guide
For use with pages 305–312
GOAL
Find probabilities.
VOCABULARY
The possible results of an experiment are outcomes. An event is an
outcome or a collection of outcomes. Once you specify an event, the
outcomes for that event are called favorable outcomes.
The probability that an event occurs is a measure of the likelihood
that the event will occur.
A theoretical probability is based on knowing all of the equally likely
outcomes of an experiment. A probability that is based on repeated
trials of an experiment is called an experimental probability.
When all outcomes are equally likely, the ratio of the number of
favorable outcomes to the number of unfavorable outcomes is called
the odds in favor of an event. The ratio of the number of unfavorable
outcomes to the number of favorable outcomes is called the odds
against an event.
Lesson 6.7
EXAMPLE
1 Finding a Probability
A jar contains 15 red marbles, 16 blue marbles, 5 yellow marbles, and 10 green
marbles. You randomly choose one marble from the jar. What is the probability
that you choose a green marble?
Solution
Because there are 10 green marbles, there are 10 favorable outcomes. There are
15 ⫹ 16 ⫹ 5 ⫹ 10 ⫽ 46 possible outcomes.
Number of favorable outcomes
Number of possible outcomes
10
5
⫽ ᎏᎏ ⫽ ᎏᎏ
23
46
P(green marble) ⫽ ᎏᎏᎏᎏ
5
23
Answer: The probability that you choose a green marble ᎏᎏ.
EXAMPLE
2 Finding Experimental Probability
You surveyed 65 randomly chosen students. Of the students you surveyed, 25
went camping during summer break. Find the experimental probability that the
next randomly chosen student went camping during summer break.
Solution
25
65
5
⫽ ᎏᎏ
13
P(camping) ⫽ ᎏᎏ
Number of successes
Number of trials
Simplify.
Answer: The experimental probability that the next randomly chosen student
5
13
went camping during summer break is ᎏᎏ, or about 0.38.
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LESSON
Name
Date
6.7 Study Guide
Continued
For use with pages 305 –312
Exercises for Examples 1 and 2
1. A drawer contains 12 white socks, 4 red socks, 6 green socks, and 10 blue
socks. You randomly choose one sock from the drawer. Find the probability
that you choose a red sock.
2. You are playing darts. Out of 45 attempts, you hit the bull’s eye 18 times.
Find the experimental probability that you hit the bull’s eye on your next
attempt.
EXAMPLE
3 Using Probability to Make a Prediction
Of the last 40 table-tennis games you played against your brother, you won 16.
Use experimental probability to predict how many games of the next 20 you
will win.
Solution
(1) Find the experimental probability that you win a game.
16
40
Lesson 6.7
P(win) ⫽ ᎏᎏ ⫽ 0.4
(2) Multiply the experimental probability by the number of games you are
going to play.
0.4 p 20 ⫽ 8
Answer: If you continue to win the same fraction of table-tennis games against
your brother, you will win 8 of the next 20 games.
EXAMPLE
4 Finding the Odds
You randomly choose a letter from a bag containing one of each letter of the
alphabet. What are the odds in favor of and the odds against choosing a vowel?
Solution
There are 5 favorable outcomes (a, e, i, o, u) and 26 ⫺ 5 ⫽ 21 unfavorable
outcomes.
Number of favorable outcomes
Number of unfavorable outcomes
5
21
Odds in favor ⫽ ᎏᎏᎏᎏ ⫽ ᎏᎏ
5
21
The odds in favor of choosing a vowel are ᎏᎏ, or 5 to 21. The odds against
21
5
choosing a vowel are ᎏᎏ, or 21 to 5.
Exercises for Examples 3 and 4
3. Of the last 25 foul shots you attempted, you made 15. Use experimental
probability to predict how many foul shots out of the next 5 you will make.
4. You spin a spinner that is divided into 20 equal parts. Six parts are orange,
8 parts are white, 3 parts are black, 2 parts are red, and 1 part is blue. Find
the odds in favor of and the odds against spinning white.
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Pre-Algebra
Chapter 6 Resource Book
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LESSON
Name
Date
6.8 Study Guide
For use with pages 313–317
GOAL
Use the counting principle to find probabilities.
VOCABULARY
One way to count the number of possibilities is to use a tree diagram.
A tree diagram uses branching to list choices.
A quick way to count the number of possibilities displayed in a tree
diagram is to use the counting principle. The counting principle uses
multiplication to find the number of possible ways two or more events
can occur. The counting principle states that if one event can occur in
m ways, and for each of these ways a second event can occur in n ways,
then the number of ways that the two events can occur together is m p n.
The counting principle can be extended to three or more events.
EXAMPLE
1 Making a Tree Diagram
You are choosing a birthday cake. It can be either round or rectangular. Vanilla,
chocolate, and lemon are your choices for cake flavors. How many different
birthday cakes are possible?
List the cake
shapes.
List the cake flavors
for each shape.
Lesson 6.8
round
rectangular
List the possibilities.
vanilla
round vanilla
chocolate
round chocolate
lemon
round lemon
vanilla
rectangular vanilla
chocolate
rectangular chocolate
lemon
rectangular lemon
Answer: Six different cakes are possible.
Exercise for Example 1
1. Suppose you also have a choice of buttercream, cream cheese, or whipped
frosting for the cake in Example 1. Copy the tree diagram above and add
the new choices. How many possible choices for a birthday cake do you
have now?
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Pre-Algebra
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LESSON
Name
Date
6.8 Study Guide
Continued
EXAMPLE
For use with pages 313–317
2 Using the Counting Principle
You are choosing a 4-character security code. The characters can be either
lowercase letters or digits. Use the counting principle to find the number of
different security codes that are possible.
There are 26 ⫹ 10 ⫽ 36 choices for each character.
36
Number of
choices for
3rd character
Number of
choices for
2nd character
Number of
choices for
1st character
p
36
p
36
Number of
choices for
4th character
p
36
Total number
of choices
⫽
1,679,616
Answer: There are 1,679,616 possible security codes.
Exercise for Example 2
2. You are choosing a computer password that has 2 uppercase letters
followed by 4 digits. Use the counting principle to find the different
passwords that are possible.
EXAMPLE
3 Finding a Probability
Your school is issuing 5-digit identification numbers to students randomly. The
first digit must not be 0. What is the probability that your identification number
is 54321?
Solution
First find the number of different identification numbers. Because the first digit
cannot be 0, there are 9 choices for the first digit and 10 choices for the other
four digits.
Lesson 6.8
9 p 10 p 10 p 10 p 10 ⫽ 90,000
Use the counting principle.
Then find the probability that your identification number is 54321.
There is only one identification
1
P(54321) ⫽ ᎏᎏ
90,000
number that is 54321.
1
90,000
Answer: The probability that your identification number is 54321 is ᎏᎏ.
Exercises for Example 3
3. You randomly choose a 7-digit code to open your garage door. What is the
probability that you choose the code 1234567?
4. You flip a coin, roll a 6-sided number cube, and spin a spinner. The spinner
is divided into 8 equal parts, labeled a, b, c, d, e, f, g, and h. Find the
probability that you flip heads, roll an even number, and spin a vowel.
Copyright © McDougal Littell/Houghton Mifflin Company
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Chapter 6
Pre-Algebra
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