answers to all exercises - weeblymissbuckles.weebly.com/uploads/2/2/4/6/22466700/daa… · ·...

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114 ANSWERS TO ALL EXERCISES

CHAPTER 10 • CHAPTER CHAPTER10 • CHAPTER 10REFRESHING YOUR SKILLS FOR CHAPTER 10

1a. 5 ___ 10 � 1 __ 2 � 0.5

1b. 6 ___ 10 � 3 __ 5 � 0.6

1c. 4 ___ 10 � 2 __ 5 � 0.4

2a. 10 ___ 36 � 5 ___ 18 � 0.2 _

7

2b. 7 is most likely; probability of 7 is 6 __ 36 � 1 _ 6 � 0.1 _

6 .

2c. 18 ___ 36 � 1 __ 2 � 0.5

3a. 26 ___ 52 � 1 __ 2 � 0.5

3b. 8 ___ 52 � 2 ___ 13 � 0.154

3c. 2 ___ 52 � 1 ___ 26 � 0.038

4a. 155 ___ 600 � 31 ___ 120 � 0.258

4b. 275 ___ 600 � 11 ___ 24 � 0.458

4c. 500 ___ 600 � 5 __ 6 � 0.8 _

3

Answers to All Exercises

DAA2TE_985_ANS_c.indd 114DAA2TE_985_ANS_c.indd 114 3/12/09 9:29:45 PM3/12/09 9:29:45 PM

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ANSWERS TO ALL EXERCISES 115

LESSON 10.1

1a. 6 ___ 15 � 0.4; 7 ___ 15 � 0.4 _

6 ; 2 ___ 15 � 0.1 _ 3

1b. experimental

2a. 698 ____ 1424 � 0.490 2b. 477 ____ 1424 � 0.335

2c. 228 ___ 435 � 0.524

2d. 263 ___ 726 � 0.362

2e. theoretical

3a. 4 ___ 14 � 0.286

3b. 10 ___ 14 � 0.714

3c. 7.5 ___ 14 � 0.536

3d. 1.5 ___ 14 � 0.107

3e. 2 ___ 14 � 0.143

4a. 5 ___ 30 � 0.1 _ 6

4b. 97.5 ____ 100 � 0.975 4c. 31 ___ 36 � 0.86 _ 1

5a. experimental

5b. theoretical 5c. experimental

6a. Answers will vary.

6b. Possible answer: Use the random integer com mand on the calculator to simulate rolling a die.

6c. Answers will vary.

6d. Answers will vary. Sum your answers from 6c and divide the answer by 10.

6e. Answers will vary. Long-run averages should tend toward 6 turns in order to roll a 6.

7. Answers will vary. Each of these methods has shortcomings.

7. i. Middle numbers (3–7) are more common than getting only 1 or 2 or 8 or 9 heads in one trial of dropping pennies.

7. ii. Very few pencils will be at 0 or 1 in.; students throw away their pencils long before that.

7. iii. This is the best method, although books tend to open to pages that are used more than others.


8b. The long-run experimental probability should show that 1 _ 6 of all rolls are a 3.

8c. Answers will vary. The points should level out to a straight line at y � 0.1

_ 6 . If you considered 5’s

instead of 3’s, the data should level out to the same value.

8d. Answers will vary but should be close to 1 _ 6 .

8e. P(3) � 1 _ 6 � 0.1 _

6 . There are six equally likely outcomes, and 3 is one of them, so the theoretical probability is 1 _ 6 .

9a. 4; 4 ___ 36 � 0. _

1

9b. 5; 5 ___ 36 � 0.13 _

8

9c. 10; 10 ___ 36 � 0.2 _

7

9d. 2; 2 ___ 36 � 0.0 _

5

9e. 10; 10 ___ 36 � 0.2 _

7

10a. 3 __ 5 , or 0.6 10b. 2 to 5

11a. 144 square units

11b. 44 square units

11c. 44 ___ 144 11d. 44 ___ 144 � 0.30 _

5

11e. 100 ___ 144 � 0.69 _

4

11f. 0; 0

12a. x � y � 6

12b. y

8

x

4

40 8

12c. 18 ___ 64 � 0.281

13a. 270

13b. 1380

13c. 270 ____ 1380 � 0.196

13d. 1110 ____ 1380 � 0.804

14a. 6 ___ 27 � 0. _

2 14b. 12 ___ 27 � 0. _

4

14c. 8 ___ 27 � 0. ___

296 14d. 1 ___ 27 � 0. ___

037

15a. 53 pm, at point C

15b. 0 pm, at point A, the nucleus


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15c. The probability starts at 0 at the nucleus, increases and peaks at a distance of 53 pm, and then decreases quickly, then more slowly, but never reaches 0.

16. x4 � 4x3y � 6x2y2 � 4xy3 � y4

17. log � ac2 ___

b �

18. x � 5000

19a. y

(–6, –1)

�5x � 4y � 26 3x � y � 15

x � 6y � �12

(2, 9)

(6, –3)

8

12

4

–4

AP1

B

C

x–4 4 8

19b. (2, 9), (�6, �1), (6, �3)

19c. 68 units2

20. a parabola with focus (3, 0) and directrix y � 6; y � � 1 __ 12 x2 + 1 _ 2 x + 9 _ 4

21a. Set i should have a larger standard deviation because the values are more spread out.

21b. i. __

x � 35, s � 22.3

21b. ii. __

x � 117, s � 3.5

21c. The original values of _

x and s are multiplied by 10.

21c. i. __

x � 350, s � 223.5

21c. ii. __

x � 1170, s � 35.4

21d. The original values of _

x are increased by 10, and the original values of s are unchanged.

21d. i. __

x � 45, s � 22.3

21d. ii. __

x � 127, s � 3.5


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LESSON 10.2

1. v1

v2

v1

v2

v1

v2

d1

d2d1

d2d1

d2d1

d2d1

d2d1

d2

e1

e2

e3

2. P(a) � 0.675; P(b) � 0.075; P(c) � 0.05; P(d) � 0.2; 1

3. P(a) � 0.7; P(b) � 0.3; P(c) � 0.18; P(d) � 0.4; P(e) � 0.8; P(f) � 0.2; P(g) � 0.08

4a. 1 __ 8 � 0.125

4b. 3 __ 8 � 0.375

4c. 2 __ 3 � 0. _

6

5a. The probability of selecting a junior given that a sophomore has already been selected; P(J2|S1)

5b. 13 ___ 20 � 0.6

5c. 7 ___ 10 � 0.7

6a. 182 ___ 420 � 0.4 _

3 ; 98 ___ 420 � 0.2 _

3 ; 98 ___ 420 � 0.2 _

3 ; 42 ___ 420 � 0.1

6b. No, because the probabilities of the four paths are not all the same

6c. 420 ____ 420 � 1

7a. 24

7b. 0.25

7c. 2 ___ 24 � 0.08 _

3

7d. 1 ___ 24 � 0.042

7e. 23 ___ 24 � 0.958

7f. 12 ___ 24 � 0.5

8a. H HH

HT

TH

TT

T

H

T

H

T

8b. HHH

HHTHTH

H

H

H

TH

TH

TH

T

T

H

T

T

HTTTHH

THTTTH

TTT

8c. HHHHHHHTHHTHHHTTHTHHHTHTHTTHHTTTTHHHTHHTTHTHTHTTTTHHTTHTTTTHTTTT

H

H

HH

TH

TH

TH

TH

TH

TH

TH

T

T

H

T

H

T

H

T

T

H

T

T

9a. 4

9b. 8

9c. 16

9d. 32

9e. 1024

9f. 2n

10a. 1 __ 2

10b. 1 __ 2

10c. independent

10d. Both statements reveal misconceptions about the probability of independent events. The probability of heads is always 1 _ 2 ; the coin does not remember how it landed previously.

11a. 1 ___ 16 � 0.0625

11b. 4 ___ 16 � 0.25

11c. 6 ___ 16 � 0.375

11d. 4 ___ 16 � 0.25

11e. 1 ___ 16 � 0.0625

11f. 1

11g. 0.3125


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12a. 0.05

0.95

0.08

0.92

0.07

0.93

M1

M2

M3

D 0.01

G 0.19

D 0.028

G 0.322

D 0.0315

G 0.4185

0.20

0.35

0.45

12b. 0.08

12c. 0.0695

12d. � 0.403

13a. 0.0289

13b. 0.9711

13c. 0.6889

13d. 0.9711

14. 6 ___ 16 � 0.375

15a. the probability that the roll is odd and it is a 3 or a 5; 1 _ 3 � 0.

_ 3

15b. the probability that the roll is a 3 or a 5 given that the roll is odd; 2 _ 3 � 0.

_ 6

15c. the probability that the roll is odd given that it is a 3 or a 5; 1

16a. 349 ____ 798 � 0.437

16b. 512 _____ 1424 � 0.360

16c. The events are dependent, because P(10th grade | female) � P(10th grade). The probability of choosing a 10th grader from the female students is greater than the probability of choosing a 10th grade student.

17. 64

18a. �3 � 2i

18b. 2 � 24i

18c. 18 ___ 29 � 16 ___ 29 i

19. P(orange) � 0.152; P(blue) � 0.45 _

6

20a. 50 ____ 110 � 0. __

45

20b. 120 ____ 230 � 0.522

21. 8 � __

2


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LESSON 10.3

1. 10% of the students are sophomores and not in advanced algebra. 15% are sophomores in advanced algebra. 12% are in advanced algebra but are not sophomores. 63% are neither sophomores nor in advanced algebra.

2a. 0.25 2b. 0.12

2c. 0.15 __________ 0.15 � 0.12 � 0. _

5

2d. 0.37

3.

50 75 60

315

Sophomore In advanced algebra

4. No. P(S) � P(A) � 0.25 � 0.27 � 0.0675, P(S and A) � 0.15. These must be equal if the events are independent.

5a. yes, because they do not overlap

5b. No. P(A and B) � 0. This would be the same as

P(A) � P(B) if they were independent.

6a.

71 55 220

74

French Music

6b. approximately 13% 6c. 74

7a.

0.12 0.08 0.32

0.48

A B

7b. i 0.08 7b. ii 0.60 7b. iii 0.48

8. 0 � P(A and B) � 0.4, 0.5 � P(A or B) � 0.9. The first dia gram shows P(A and B) � 0 and P(A or B) � 0.9. The second dia gram shows P(A and B) � 0.4 and P(A or B) � 0.5.

0.4 0.5

A B

0.40.1

AB

9.

0.075 0.175 0.12

0.63

Sophomore In advanced algebra

10a. yellow

10b. cyan

10c. white

10d. blue

10e. green

10f. black

11a.

0.18 0.045

0.015

0.14

0.105

0.06 0.035

Amber Bob

Carol

11b. 0.015 11c. 0.42

12a. 280 _____ 1500 � 0.18 _

6 12b. 775 _____ 1500 � 0.51 _

6

12c. 145 ____ 355 � 0.408 12d. 145 ____ 775 � 0.187

13. approximately 77

14. 324 _____ 15625 � 0.021

15a. 3 � __

2

15b. 3 � __

6

15c. 2xy2 � _____

15xy

16. 2 __ 3 � 0. _

6

17. Answers will vary. Making a free throw is not random as flipping a coin is, so Janie may be improving. However, even if her overall accuracy is still 50%, she could have a sequence of 5 successes in a row.


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LESSON 10.4

1a. Yes; the number of children will be an integer, and it is based on a random process.

1b. No; the length may be a non-integer.

1c. Yes; there will be an integer number of pieces of mail, and it is based on random processes of who sends mail when.

2a. Yes; the result of each call is independent of other calls, and you stop counting when she is successful.

2b. No; the number of cats is a discrete random variable, but you don’t stop counting when you get the first cat; it is not geometric.

2c. No; you are counting minu tes until you hear a song, but because not all songs are the same length and minutes are not equivalent to songs, you are working with two different types of variables.

3a.

x 0 1 2 3

P(x) 1 __ 8 3 __ 8 3 __ 8 1 __ 8

3b. 1.5

4a. approximately 0.068

4b. approximately 0.203

5a. 0

5b. 0

6a. Answers will vary. Theoreti cally, after 10 games Sly should get about 23 points, and Andy should get about 21.

6b. Answers will vary. Theoreti cally, it should be close to 0.41.

6c. Andy gets 5.

Sly gets 4.

15__36

21__36

6d. �0.25

6e. Answers will vary. One possible answer is 5 points for Sly if the sum of the dice is less than 8 and 7 points for Andy if the sum of the dice is greater than 7.

7a. $25

7b. 0. _ 6

7c. approximately $28.33


8b. Sample answer: Assign each of the letters in the word CHAMPION a different number from 1 to 8. Randomly generate numbers between 1 and 8. Count how many digits you must gene rate until you have at least one of each number.

8c. Answers will vary.

8d. Answers will vary. Theoretically, it should be about 22 boxes.

8e. Answers will vary. The average number of boxes should be about 22.

9a. 0.2

9b. 0.83 � 0.2 = 0.1024

9c.

Successful hits Probability

0 0.2

1 0.16

2 0.128

3 0.1024

4 0.08192

5 0.065536

9d. P(n) = 0.2(0.8)n

9e. geometric

9f. P(n � 6) � 1 � � i�0

6

0.2 (0.8)i � 0.210

10a. 6. _ 8 � 7 points

10b. Answers will vary.

11a. 0.580

11b.

Number of defective radiosx

ProbabilityP(x) x � P(x)

0 0.420 0

1 0.312 0.312

2 0.173 0.346

3 0.064 0.192

4 0.031 0.124

5 0.000 0

11c. 0.974


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11d. On average, the engineer should expect to find 0.974 defective radio in a sample of 5.

12. approximately 0.047

13. 1

14. 0.4

15a.

0.04 0.16

0.33

0.07 0.03

0.07

0.13 0.17

Metal Oval

Small

15b. Calculated using the actual frequencies:

Metal

Plas.Metal

0.69 Small

Oval

Tri.

Tri.

Oval

Shape

Size

Material

Large

0.72

0.65

0.350.64

0.360.70

0.300.55

0.45

0.28

0.75

0.25

0.31

Plas.Metal

Plas.Metal

Plas.

16. n �5 ______ log � 1 _ 2 �

� 16.61

17. 44


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LESSON 10.5

1a. Yes. Different arrangements of scoops are different.

1b. No. We are not counting different arrangements separately.

1c. No. Repetition is not allowed in permutations.

1d. No. Repetition is not allowed in permutations.

2a. 12

2b. 7

2c. n � 1

2d. n

2e. 14,280

2f. n(n � 1)

2g. n � 14

3a. 210

3b. 5040

3c. (n � 2)!

_______ 2

3d. n! __ 2

4a. 24

4b. 18

4c. 256

4d. 192

5a. 10,000; 27. _

7 h

5b. 100,000; approximately 11.57 d

5c. 10

6. n � r � 6, or n � 6 and r � 5, or n � 10 and r � 3, or n � 720 and r � 1

7. r factors

8. 60

9a. 40,320

9b. 5040

9c. 0.125

9d. Sample answer: There are eight possible positions for Volume 5, all equally likely. So P(5 in rightmost slot) � 1 _ 8 � 0.125.

9e. 0.5; sample answer: there are four even-numbered books that can be in the rightmost position out of the eight books. So the probability of an even-numbered book being on the right is 4 _ 8 � 0.5.

9f. 1

9g. 40,319

9h. 1 ______ 40,320 � 0.000025

10.

11a. 100,000

11b. 1,000,000,000

11c. 17,576,000

11d. 7,200,000

12a. 1.516 � 1016

12b. 2000 min � 33.3 h, or about 1 d 9 h 20 min

12c. about 3.639 � 1017 min, or about 6.92 � 10 9 centuries

13a. 0

13b. 1 __ 4 � 0.25

13c. 2 __ 4 � 0.5

13d. It is not possible because there are no brown-eyed (B) genes in the mixture.

13e. 4 ___ 16 � 0.25

14a. 30 ___ 50 � 0.6

14b. 16 ___ 30 � 0.5 _

3

15a. �0.5

15b. You would expect to lose 0.50 point on each of the 10 tosses, or a total loss of 5 points.

16a. 1 __ 8 � 0.125

16b. 3 __ 8 � 0.375

16c. 1 __ 2 � 0.5

17a. y � �0.25x2 � 2.5x � 3.25

17b. y � �0.25(x � 5)2 � 3

17c. y � �0.25�x � 5 � 2 � __

3 � ��x � 5 � 2 �

__ 3 �

18a. 41

18b. about 808.3 in2

NNumber of

permutations of N items Time

5 120 0.00012 s

10 3,628,800 3.6288 s

12 479,001,600 � 8 min

13 6,227,020,800 � 1.7 h

15 � 1.31 � 1012 � 15 d

20 � 2.43 � 1018 � 77,100 yr

NNumber of

permutations of N items Time

5 120 0.00012 s

10 3,628,800 3.6288 s

12 479,001,600 � 8 min

13 6,227,020,800 � 1.7 h

15 � 1.31 � 1012 � 15 d

20 � 2.43 � 1018 � 77,100 yr


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LESSON 10.6

1a. 120 1b. 35

1c. 105 1d. 1

2a. 120 2b. 35

2c. 105 2d. 1

3a. 7P2 ___ 2!

� 7C2

3b. 7P3 ___ 3!

� 7C3

3c. 7P4 ___ 4!

� 7C4

3d. 7P7 ___ 7!

� 7C7

3e. nPr ___ r!

� nCr

4. Neither; they are the same.

5. n � 7 and r � 3, or n � 7 and r � 4, or n � 35 and r � 1, or n � 35 and r � 34

6. r � 6; 10! ___ 4! 6! � 10!

___ 6! 4! ; the number of 10 things taken 4 at a time is equal to the number of 10 things omitted 6 at a time.

7a. 35

7b. 20 ___ 35 � 0.571

8. Sample answer: In a true “combination” lock, the order in which the numbers are en tered would not matter. In com bination locks, the order of the numbers does matter, so they are more like permutation locks. However, in a true “permutation” lock, repeated numbers would not be allowed.

9a. 4 9b. 8 9c. 16

9d. The sum of all possible combinations of n things is 2n; 25 � 32.

10a. approximately 3.4 yr

10b. 1000 _____ 47C6

� 0.000093

11a. 6

11b. 10

11c. 36

11d. nC2 � n! ________ 2(n � 2)!

12a. 26,466,926,850 ways

12b. approximately 2.134 1019 ways

13a. 47C3 ____ 50C4

� 0.070

13b. 47C2 ____ 50C4

� 0.005

13c. 1 � 48C4 ____ 50C4

� 0.155

13d. $3.20

14a. x2 � 2xy � y2

14b. x3 � 3x2y � 3xy2 � y3

14c. x4 � 4x3y � 6x2y2 � 4xy3 � y4

15. 15 speeds

16a. 0.0194 is the probability that someone is healthy but tests positive.

16b. 0.02 is the probability that a healthy person tests positive.

16c. 0.0491 is the probability that a person tests positive.

16d. 0.395 is the probability that a person who tests positive is healthy.

17. C � 157 ____ 4 , or C � 39.25

18a. $26,376.31

18b. 20 yr 11 mo

19a. � 1 ____ �

__ 2 , 1 ____

� __

2 � , or � �

__ 2 ____ 2 , �

__ 2 ____ 2 �

19b. � 1 __ 2 , � __

3 ____ 2 �


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LESSON 10.7

1a. x 47

1b. 5,178,066,751x 37y 10

1c. 62,891,499x 7y 40

1d. 47x y 46

2a. 0.75

2b. 0.0625

2c. (0.25)n

2d. approximately 0.264

3a. 0.299

3b. 0.795, 0.496

3c. 0.203, 0.502, 0.791

3d. Both the “at most” and “at least” numbers include the case of “exactly.” For example, if “exactly” 5 birds (0.165) is subtracted from “at least” 5 birds (0.203), the result (0.038) is the same as 1 � 0.962 (“at most” 5 birds).

3e. The probability that at least 5 birds survive is 20.3%.

4. What is the probability of exactly 35 successes in 50 trials?

5. What is the probability of at least 35 successes in 50 trials?

6a. HH, HT, TH, TT

6b. HH, HT, TH, TT

6c. Both diagrams would look the same.

H HH

HT

TH

TT

T

H

T

H

Second flip/coin

First flip/coin

T

6d. 2C 0 � 1 is the number of ways of getting 0 tails,

2C 1 � 2 is the number of ways of gett ing 1 tail, and

2C2 � 1 is the number of ways of getting 2 tails.

6e. There is 1 way of getting 2 heads; there are 2 ways of getting 1 head and 1 tail; and there is 1 way of getting 2 tails.

7a. x 4 � 4x 3y � 6x 2y 2 � 4xy 3 � y 4

7b. p 5 � 5p 4q � 10p3q2 � 10p 2q 3 � 5pq 4 � q 5

7c. 8x 3 � 36x 2 � 54x � 27

7d. 81x 4 � 432x 3 � 864x 2 � 768x � 256

8a. 50C40 � p 10q 40 or 50C10 � p 10q 40

8b. the sum of terms 1 to 11, or � r�0

10

50Cr p r (1 � p)50�r

8c. P(r � 10) � 1.193 � 10�5

8d. no

9a. approximately 0.401

9b. approximately 0.940

9c. f (x) � 30Cx(0.97)30�x(0.03)x

9d. 0.940

10. 0.007

11a. 0.000257

11b. 0.446

11c. 0.983

12a. 0.049

12b. 0.079

12c. 1.88 birds, or approximately 2 birds

13. Answers will vary. This event will happen in 15.625% of trials.

14a. 0.025

14b. 0.004

14c. 0.0002

14d. 0.0288

15a.

x 1 2 3 4

Sum of the first 2 terms

2 2 2 2

Sum of all the terms 2 2.25 � 2.370 � 2.441

15b. f (10) � 2.594, f (100) � 2.705, f (1,000) � 2.717, f (10,000) � 2.718

15c. There is a long-run value of about 2.718.

16. 65,780. Sample answer: Either the group of five students selected includes the new student or it doesn’t. If the new student is included, then the other 4 are selected from the remaining 25 class members, and this can be done 25C4 � 12,650 ways. If the new student is not selected, then all 5 are selected from the 25 original members, and this can be done 25C5 � 53,130 ways. This means there are 12,650 ways that the new student is part of the group and 53,130 ways that he or she is not. This makes 12,650 � 53,130 � 65,780 ways to select 5 students.

17a. The experimental pro babilities are likely to be differ ent from 0.5 and 0.5. In this sample simulation, P(H) � 0.6 and P(T) � 0.4.


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17b. The experimental probabilities should be closer to 0.5 and 0.5, but are likely to not yet be exact. In this sample simulation, P(H) � 0.49 and P(T) � 0.51.

17c. You can model the thumbtack by defining randomPick with more “points up” to choose from. For example, randomPick(“U”,“U”,“U”,“D”, “D”) would make the experimental probability of “points up” approximately 0.6.

18. 37.44 cm2

19a. (distance, period)

x

y

9000

12,000

Distance (millions of miles)

Per

iod

(log (distance), period)

x

y

87 9

12,000

0

Log distance

Per

iod

(distance, log (period))

x

y

900

1

0

2

3

4

Distance (millions of miles)

Log

per

iod

(log (distance), log (period))

x

y

87 9

1

0

2

3

4

Log distance

Log

per

iod

(log (distance), log (period)) is the most linear of the four, so use the median-median line to find an equation to fit these points; y � 1.50x � 9.38; log (period) � 1.50 log (distance) � 9.38;

10 log (period) � 10 1.50 log (distance) �9.38;period � distance 1.50 10 �9.38

19b. 31,385; 61,566; 92,535; errors may be due to rounding. For more accurate results, the a- and b-values found from regression can be stored in variables in the calculator.

19c. period 2 � 10�18.76 distance3

20a. A � � r2, where A � area and r � radius

20b. V � l � w � h, where l � length, w � width, and h � height

20c. N � k _ d , where d � the distance from the hinge


An

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CHAPTER 10 REVIEW

1. Answers will vary depending upon whether you interpret the problem to imply random decimal numbers (between 0 and 10, non-inclusive) or random integers (0 to 10, inclusive). To generate random decimal numbers, you might look at a random-number table and place the decimal point after the first digit in each group of numbers. Alternatively, you could use a calculator command, such as 10*rand(numTrials) on the TI-Nspire, or 10*rand on the TI-84 Plus. To generate random integers, you might number 11 chips or slips of paper and randomly select one. Alternatively, you could use a calculator command, such as randInt(0, 10) on the TI-Nspire or the TI-84 Plus.

2a, b. 8

6

7

5

4

3

2

1

2 3 4 51 876

2c. 10 ___ 64 � 0.15625

2d. 49 ___ 64 � 0.766

3a. 0.5

3b. 17.765 units2

4a. TTTT

TTTFTTFT

TTFF

T

FT

F T

F

T

F

T

F

T

F

T

F

TFTT

TFTFTFFT

TFFFFTTT

FTTFFTFT

FTFFFFTT

FFTFFFFT

FFFF

TF

TF

TF

TF

TF

TF

TF

TF

4b. 4

4c. Because the order in which the true and false answers occur doesn’t matter, use combinations:

4C3 � 4.

4d. 2 __ 4 � 0.5

5a.

Plain0.36

P 0.62

R 0.11M 0.27

P 0.62

R 0.11M 0.27

P 0.62

R 0.11M 0.27

Veggie0.17

Chili0.47

5b. 0.0517

5c. 0.8946 5d. 0.3501

6a. See below.

6b. 37 ___ 55 � 0.6 __

72 6c. 37 ____ 122 � 0.303

6d. 18 ____ 254 � 0.071 6e. 118 ____ 372 � 0.317

7. 110.5

8.

0.16 0

0

0.04 0.42

0.20

0.06 0.12

Cats Dogs

Other

9. approximately 0.044

10a. 1

10b. x99 ____

1299

10c. 293, 930a12b9

9th grade 10th grade 11th grade 12th grade Total

Ice cream 18 37 85 114 254

Whipped cream 5 18 37 58 118

Total 23 55 122 172 372

6a. (Chapter 10 Review)


answers to all exercises - weeblymissbuckles.weebly.com/uploads/2/2/4/6/22466700/daa… · ·...

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