Math 141 Exam 2 Review Answer Key

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Math 141 WIR, Spring 2007, c ... Math 141 Exam 2 Review Answer Key ... 2. (b) 3. 4. (c) 4. 13. (d) 9. 13. 7. (a) P(20,20) = 20! (b) 3! · 9! · 6! · 5! 8. P(200,2) · C(198 ...
c Math 141 WIR, Spring 2007, Benjamin Aurispa

Math 141 Exam 2 Review Answer Key 1. (a) S = {(H, R), (H, B), (H, G), (T, R), (T, B), (T, G)} (b) E = {(H, R), (T, R)} (c) F = {(T, R), (T, B), (T, G), (H, G)} (d) No, E ∩ F = {(T, R)} (e) No, all outcomes are not equally likely. There are not an equal number of marbles of the different colors. 2. (a) Yes, all outcomes are equally likely. (b) E = {(1, 6), (2, 3), (3, 2), (6, 1), (2, 4), (4, 2)} P (E) = 16 (c) F = {(1, 4), (4, 1), (2, 2), (1, 3), (3, 1)} 5 P (F ) = 36 3. (a)

Simple Event Probability

Y

13 44 11 9 + 44 44

G

O

R

P

11 44

9 44

6 44

5 44

20 44

or

5 11

(c) P (P c ) = 1 − P (P ) = 1 −

5 44

=

(b) P (G ∪ O) =

=

39 44

4. Note: I had changed some of the numbers for this problem to P (E) = .4, P (F ) = .5, and P (E ∪ F ) = .6 so that it would actually make sense. These changes have been made on the problem sets and are also reflected in the full solutions. (a) 0.3 (b) 0.5 (c) 0.7 (d) 0.2 5. (a) (b) (c) (d) 6. (a) (b) (c) (d)

89 500 175 500 350 500 325 500

= 0.178 = 0.35 = 0.7 = 0.65

1 2 3 4 4 13 9 13

7. (a) P (20, 20) = 20! (b) 3! · 9! · 6! · 5! 8. P (200, 2) · C(198, 18) 1

c Math 141 WIR, Spring 2007, Benjamin Aurispa

9. (a) C(7, 4) · C(19, 2) = 5, 985 (b) Method 1: C(9, 3)C(17, 3) + C(9, 4)C(17, 2) + C(9, 5)C(17, 1) + C(9, 6) = 76, 482 Method 2: C(26, 6) − [C(17, 6) + C(9, 1)C(17, 5) + C(9, 2)C(17, 4)] = 76, 482 (c) C(17, 6) = 12, 376 (d) C(10, 3)C(16, 3) + C(7, 2)C(19, 4) − C(10, 3)C(7, 2)C(9, 1) = 125, 916 10.

18! 7!·6!·5!

11. (a) True (b) False (c) False (d) True (e) False (f) True (g) False (h) True 12. (a) 32 (b) 85 (c) 69 13. 5 · 5 · 4 · 4 · 3 = 1200 14. See full solutions from the live review for graph. Corner Points (2, 8) (2, 6) (6, 4)

C = 2x + 4y 36 28 28

C is minimized at infinitely many points which all lie on the line segment between (2, 6) and (6, 4). The minimum value of C is 28. 15. Note: I changed the problem to 35 ounces of jelly available instead of 24. This change is reflected on the problem sets and on the full solutions from the live review. x = number of small sandwiches sold each hour y = the number of large sandwiches sold each hour Maximize P = 2x + 5y Subject to: 3x + 5y ≤ 30 4x + 5y ≤ 35 x≥0 y≥0

2

c Math 141 WIR, Spring 2007, Benjamin Aurispa

Corner Points (0, 0) (0, 6) (8.75, 0) (5, 3)

P = 2x + 5y 0 30 17.5 25

Profit is maximized when 0 small and 6 large sandwiches are sold. The maximum profit is $30. There are no ounces of peanut butter left over, but there are 5 ounces of jelly left over. 16. x = number of children tickets sold y = number of student tickets sold z = number of adult tickets sold Maximize P = 10x + 20y + 25z Subject to: x + y + z ≤ 15000 z ≥ 4x 20y ≤ 13 (25z) z ≥ 12 (x + y + z) x≥0 y≥0 z≥0

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