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Chapter 0 Posttest
State the domain and range of each relation. Then determine whether each relation is a function. Write
yes or no.
1. SOLUTION: The domain is the set of x-coordinates.
The domain is the set of y-coordinates.
Because 0 is paired with −2 and 12, this is not a function.
2. SOLUTION: The domain is the set of x-coordinates.
The domain is the set of y-coordinates.
Because −2 is paired with −1 and 2, and similarly 2 is paired with −1 and 3, this is not a function.
Name the quadrant in which each point is located.
3. (–3, 7)
SOLUTION: The point (–3, 7) has a negative x-coordinate and a positive y-coordinate. The point is located in Quadrant II.
4. (10, –11)
SOLUTION: The point (10, –11) has a positive x-coordinate and a negative y-coordinate. The point is located in Quadrant IV.
5. (–15, –3)
SOLUTION: The point (–15, 3) has a negative x-coordinate and a negative y-coordinate. The point is located in Quadrant III.
Find each product.
6. SOLUTION: Use the FOIL method to find the product.
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The point (10, –11) has a positive x-coordinate and a negative y-coordinate. The point is located in Quadrant IV.
5. (–15, –3)
SOLUTION: Chapter
0 Posttest
The point (–15, 3) has a negative x-coordinate and a negative y-coordinate. The point is located in Quadrant III.
Find each product.
6. SOLUTION: Use the FOIL method to find the product.
7. SOLUTION: Use the FOIL method to find the product.
8. SOLUTION: 9. SOLUTION: Use the FOIL method to find the product.
10. GEOMETRY The height of a rectangle is 3 millimeters less than twice the width.
a. Write an expression for each measure.
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b. Write
a polynomial expression for the area of the rectangle.
SOLUTION: Page 2
Chapter 0 Posttest
10. GEOMETRY The height of a rectangle is 3 millimeters less than twice the width.
a. Write an expression for each measure.
b. Write a polynomial expression for the area of the rectangle.
SOLUTION: a. Let w be the width of the rectangle.
The height of the rectangle is 3 millimeters less than twice the width, or (2w − 3).
b. The area of the rectangle is the product of the expressions for the width and height.
Factor each polynomial completely.
11. SOLUTION: This is a perfect square trinomial.
12. SOLUTION: This is a perfect square trinomial.
13. SOLUTION: This is a perfect square trinomial.
14. SOLUTION: This can be factored as a difference of squares.
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Chapter 0 Posttest
14. SOLUTION: This can be factored as a difference of squares.
15. STUDENT COUNCIL A student council has 6 seniors, 5 juniors, and 1 sophomore as members. In how many
ways can a 3-member council committee be formed that includes one member from each class?
SOLUTION: 6 ∙ 5 ∙ 1 = 30
A 3-member council committee can be formed in 30 ways.
Determine whether each situation involves permutations or combinations. Then solve.
16. How many ways are there to select one competitor and one alternate out of 8 students? SOLUTION: When selecting competitor and an alternate, the order does matter. Therefore this is a permutation.
nPr = =
= 56 different ways
17. How many ways are there to form a team of 7 athletes from a group of 15 who try out?
SOLUTION: When forming teams, order does not matter. Therefore this is a combination.
nCr = =
=
= 6435 ways
RESTAURANT There are 24 male and 36 female patrons in a restaurant. Of the 11 patrons under 10
years old, 6 are male. Of the 14 patrons over 55 years old, 9 are female. A patron is selected at random.
Determine whether the events are mutually exclusive or not mutually exclusive. Then find each
probability.
18. P(female or under 10)
SOLUTION: To be mutually exclusive, two events can not occur at the same time. It is possible to be both female and under 5. So
P(female or < 5) is not mutually exclusive.
P(female or < 10) =
or
19. P(under 10 or over 55)
SOLUTION: To be mutually exclusive, two events can not occur at the same time. It is not possible for a person to be both under
10 and over 55.. So P(<10 or > 55) is mutually exclusive.
P(<10
or >- Powered
55) = by Cognero
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or
Determine whether the events are independent or dependent. Then find the probability.
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SOLUTION: To be mutually exclusive, two events can not occur at the same time. It is possible to be both female and under 5. So
P(female or < 5) is not mutually exclusive.
Chapter
0 Posttest
P(female
or < 10) =
or 19. P(under 10 or over 55)
SOLUTION: To be mutually exclusive, two events can not occur at the same time. It is not possible for a person to be both under
10 and over 55.. So P(<10 or > 55) is mutually exclusive.
P(<10 or > 55) =
or
Determine whether the events are independent or dependent. Then find the probability.
20. Slips of paper numbered 1 through 10 are placed into a bag. What is the probability of drawing the number 10 three
times in a row if a slip is drawn at random and then replaced?
SOLUTION: Since the paper is replaced each time, the probability of each turn is independent. The roll does not affect the
probability of the toss P(10, 10, and 10) = or
.
21. Two students are selected at random from a class that consists of 13 males and 7 females. What is the probability
that both students are female?
SOLUTION: Selecting two students with our replacing is a dependent event. The student picked first effects the probability of the
student chosen next. P(female then female) =
or
.
22. TESTING Of the students who took both the Mid-Chapter 4 Quiz and the Chapter 4 Test, 56% passed the quiz and
48% passed both the quiz and the test. If a student passed the quiz, find the probability that he or she also passed the
test.
SOLUTION: 23. Determine whether the rectangles are similar, congruent, or neither.
SOLUTION: Since the sides of the rectangles are neither congruent nor proportional, the rectangles are neither congruent nor
similar.
24. COMPUTERS A computer image of a painting is 320 pixels wide by 240 pixels high. If the actual painting is 42
inches wide, how high is it?
SOLUTION: Let x be the height of the actual painting.
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SOLUTION: Since0the
sides of the rectangles are neither congruent nor proportional, the rectangles are neither congruent nor
Chapter
Posttest
similar.
24. COMPUTERS A computer image of a painting is 320 pixels wide by 240 pixels high. If the actual painting is 42
inches wide, how high is it?
SOLUTION: Let x be the height of the actual painting.
The height of the actual painting is 31.5 inches.
Find each missing measure. Round to the nearest tenth, if necessary.
25. SOLUTION: The missing measure of the leg of the right triangle is 24 inches.
26. a = 33 cm, b = ? cm, c = 45 cm
SOLUTION: The missing measure of the leg of the right triangle is about 30.6 cm.
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The lengths of three sides of a triangle are given. Determine whether each triangle is a right triangle.
Chapter
0 Posttest
The missing
measure of the leg of the right triangle is 24 inches.
26. a = 33 cm, b = ? cm, c = 45 cm
SOLUTION: The missing measure of the leg of the right triangle is about 30.6 cm.
The lengths of three sides of a triangle are given. Determine whether each triangle is a right triangle.
27. 6 in., 8 in., 12 in.
SOLUTION: Because the longest side is 12 in., use 12 as c, the measure of the hypotenuse.
2
2
2
Because c ≠ a + b , the triangle is not a right triangle.
28. 30 m, 34 m, 16 m
SOLUTION: Because the longest side is 34 m, use 34 as c, the measure of the hypotenuse.
2
2
2
Because c = a + b , the triangle is a right triangle.
Find the mean, median, mode, range, and standard deviation of each data set. Then identify any outliers.
29. number of students present at 8 student council meetings: 23, 45, 16, 75, 32, 35, 28, 35
SOLUTION: Enter the data into L1.
Press
to display the 1-Var statistics.
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Chapter 0 Posttest
2
2
2
Because c = a + b , the triangle is a right triangle.
Find the mean, median, mode, range, and standard deviation of each data set. Then identify any outliers.
29. number of students present at 8 student council meetings: 23, 45, 16, 75, 32, 35, 28, 35
SOLUTION: Enter the data into L1.
Press
to display the 1-Var statistics.
The mean is 36.1 students.
The median is 33.5 students.
The mode is 35 students.
The range is 75 – 16 or 59 students.
The standard deviation is 16.8 students.
To identify outliers, you need the Q1 and Q3.
Q1 = 25.5 and Q3 = 40. IQR = Q3 – Q1 or 40 – 25.5 = 14.5.
Find and use the IQR to find the values beyond which any outlier would lie.
Q1 – 1.5(IQR) and Q3 + 1.5(IQR) 25.5 – 1.5(14.5) 40 + 1.5(14.5) 3.75 61.75
The interval beyond which any outliers would lie is 3.75 < x < 61.75. Since 75 > 61.75, it is an outlier.
30. running time in minutes for 17 movies: 95, 102, 148, 140, 110, 103, 107, 104, 99, 111, 109, 124, 109, 90, 92, 110, 129
SOLUTION: Enter the data into L1.
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Press
to display
the 1-Var statistics.
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25.5 – 1.5(14.5) 40 + 1.5(14.5) 3.75 61.75
Chapter
0 Posttest
The interval
beyond which any outliers would lie is 3.75 < x < 61.75. Since 75 > 61.75, it is an outlier.
30. running time in minutes for 17 movies: 95, 102, 148, 140, 110, 103, 107, 104, 99, 111, 109, 124, 109, 90, 92, 110, 129
SOLUTION: Enter the data into L1.
Press
to display the 1-Var statistics.
The mean is about 110.7 min.
The median is 109 min.
The mode is 109 and 110 min.
The range is 148 – 90 or 58 min.
The standard deviation is about 15.6 min.
To identify outliers, you need the Q1 and Q3.
Q1 = 100.5 and Q3 = 117.5. IQR = Q3 – Q1 or 117.5 – 100.5= 17.
Find and use the IQR to find the values beyond which any outlier would lie.
Q1 – 1.5(IQR) and Q3 + 1.5(IQR) 100.5 – 1.5(17) 117.5 + 1.5(17) 75 143
The interval beyond which any outliers would lie is 75 < x < 143. 148 min is an outlier. eSolutions Manual - Powered by Cognero
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