10-6. According to the U.S. Census Bureau, of the 312.59 million

Algebra: 10.1.1 (Day 2): Categorical Data
Solutions
Name __________________________
Block ____ Date ____________
Bell Work: Use the Zero Product Property to solve each quadratic equation.
a.
(3π‘₯ βˆ’ 5)(π‘₯ + 7) = 0
3π‘₯ βˆ’ 5 = 0
3π‘₯ = 5
πŸ“
𝒙=
πŸ‘
π‘₯+7=0
𝒙 = βˆ’πŸ•
b. 2π‘₯ 2 βˆ’ 162 = 0
π‘₯ 2 βˆ’ 81 = 0
(π‘₯ + 9)(π‘₯ βˆ’ 9) = 0
𝒙 = βˆ’πŸ— 𝒙 = πŸ—
c.
+12 36π‘₯ βˆ’12
π‘₯ 3π‘₯ 2 βˆ’1π‘₯
3π‘₯
βˆ’1
3π‘₯ 2 + 35π‘₯ βˆ’ 12 = 0
3π‘₯ βˆ’ 1 = 0
3π‘₯ = 1
𝟏
𝒙=
πŸ‘
π‘₯ + 12 = 0
𝒙 = βˆ’πŸπŸ
10-5. A researcher suspected that some blood types are more susceptible to the DapT
genetic defect than other blood types. The researcher collected the following data to
investigate. Make a relative frequency table to determine whether there is an
association between blood type and the presence of the DapT defect.
All the two-way tables in this lesson are displayed, for convenience,
with the variable across the top as the independent variable. That
means that you calculate percents for columns. The percents in a
column should add up to 100%.
10-6. According to the U.S. Census Bureau, of the 312.59 million people living in the U.S. in 2011, the
number of people living alone is given in the two-way table below (in millions).
a. Why do you think this is called a two-way table?
b. What is the probability of living alone?
c. What is the probability of being 65 or over?
d. What is the probability of being under 65?
e. What is the probability of being under 35 years of
age and living alone?
f.
To find the probability of being under 35 years of age or living alone, Qui added the total number of people in the
first two columns to the total number of people in the first row. What did Qui do wrong?
10-7. Refer to problem 10-6 above. Is there an association between age and living alone?
10-8. In a recent survey of college freshman, 35% of students checked the box next to β€œExercise regularly,” 33%
checked the box next to β€œEat five or more servings of fruits and vegetables a day,” and 57% checked the box next to
β€œNeither.”
a.
Draw a two-way table to represent this situation.
b.
What is the probability that a freshman who was
surveyed exercises regularly and eats 5 servings of
fruits and vegetables each day?
c.
What is the probability that a freshman who was
surveyed exercises regularly or eats 5 servings of
fruits and vegetables each day?
d.
Is there an association between eating fruits and
vegetables and exercise? Create a relative
frequency table to help answer this question.
10-9. Is there an association between chores and curfews? In other words, do students who do more chores also
have a stricter curfew?
a.
The data for 27 students in the two-way table below is fabricated (β€œmade up”) to make a point.
Assume that curfew is the independent variable. That means that you
should find the total number of students having early curfew (or late
curfew) when you create a relative frequency table. Then the columns of
β€œcurfew” will add up to 100%.
b.
Is there an association between chores and curfew? Explain based on the relative frequency table you made.
c.
Does doing chores depend on whether you have an early or late curfew, or does having an early or late curfew
depend on whether you do chores or not? Does it matter? Let’s explore those questions. This time, assume that
chores is the independent variable. Make a new relative frequency table, by finding the totals of doing or not
doing chores. Since chores is the independent variable, we interchange the rows and columns.
Does changing the independent variable change your conclusion about there being an association?
d.
Now consider this new group of 27 fabricated students. Is there an association
if curfew is the independent variable? Is there an association if chores is the
independent variable? Make relative frequency tables for both situations.
e.
What is interesting about the relative frequency tables that have no association?
10-16. Mr. McGee is the manager of the washer and dryer department of an appliance store. He wants to estimate
the probability that a customer who comes to his department will buy a washer or dryer. He collected the following
data during one week: 177 customers came to his department, 88 purchased washers, 64 purchased dryers, and 69
did not make a purchase.
a.
Make a two-way table. Using Mr. McGee’s data, what
is the probability that the next customer who comes to
his department will purchase a washer or a dryer?
b.
Mr. McGee promises his sales people a bonus if they can
increase the probability that the customers who buy
washers also buy dryers. What is the probability that if a
customer bought a washer, he or she also bought a
dryer?
10-17. Now David wants to solve the equation 4000x – 8000 = 16,000.
a.
What easier equation could he solve instead that would
give him the same solution? (In other words, what
equivalent equation has easier numbers to work with?)
b.
Justify that your equation in part (a) is equivalent to
4000x – 8000 = 16,000 by showing that they have the
same solution.
c.
David’s last equation to solve is 100 + 100 = 100. Write
and solve an equivalent equation with easier numbers
that would give him the same answer.
π‘₯
3
8
Show process on b. and c.
10-18. Match each graph below with the correct inequality.
d.
b.
a. y > –x + 2
a.
c.
1
c. y β‰₯ 2 π‘₯
b. y < 2x – 3
d. y <
βˆ’2
3
π‘₯+2
10-19. The cost of a loaf of bread is now $2.79 and has been increasing 8% per year for several years.
a.
At the same rate, what will it cost in 5 years?
b.
What did it cost 5 years ago?
𝑦 = 2.79(1.08)5
π’š β‰ˆ $πŸ’. 𝟏𝟏
𝑦 = 2.79(1.08)βˆ’5
π’š β‰ˆ $𝟏. πŸ—πŸŽ
10-20. Solve the following inequalities for x and graph solutions on a number line.
a.
4x – 1 > 7
c.
2(x – 5) < 8
π‘₯βˆ’5≀ 4
π’™β‰€πŸ—
4π‘₯ β‰₯ 8
𝒙β‰₯𝟐
b.
3 – 2x < x + 6
d.
3 βˆ’ 3π‘₯ < 6
βˆ’3π‘₯ < 3
𝒙 > βˆ’πŸ
1
2
π‘₯>5
𝒙 > 𝟏𝟏
10-21. Factor each polynomial.
a.
64x2 – y2
(πŸ–πŸ– + π’š)(πŸ–πŸ– βˆ’ π’š)
b.
12x2 – xy – 6y2
(πŸ‘π’™ + πŸπ’š)(πŸ’π’™ βˆ’ πŸ‘π’š)
+𝑦 8π‘₯𝑦
βˆ’π‘¦ 2
8π‘₯
βˆ’π‘¦
c.
(πŸπ’™ + πŸ‘)𝟐
8π‘₯ 64π‘₯ 2 βˆ’8π‘₯𝑦
+2𝑦 8π‘₯𝑦 βˆ’6𝑦 2
3π‘₯ 12π‘₯ 2 βˆ’9π‘₯𝑦
4π‘₯
βˆ’3𝑦
4x2 + 12x + 9
d.
x2 – 3x + 18
𝑷𝑷𝑷𝑷𝑷
+3 6π‘₯
9
2π‘₯ 4π‘₯ 2 6π‘₯
2π‘₯ +3
π‘₯2
18