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zaharov [31]
3 years ago
9

Evaluate d-17 when d=30​

Mathematics
2 answers:
adelina 88 [10]3 years ago
5 0

d-17

d=30

Replace d

30-17=13

3241004551 [841]3 years ago
3 0

Answer:

d-17

= 30-17

= 13

Step-by-step explanation:

Since d = 30, so d-17= 13

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The answer is 95.

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The table below shows three unique functions.
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A common way for two people to settle a frivolous dispute is to play a game of rock-paper-scissors. In this game, each person si
Leokris [45]

Answer:

(a) P(A) = 0.34

(b) P(B) = 0.33

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(d) The Complement of event A = 1 - P(A) = 0.66

Step-by-step explanation:

We are given that in the long run, roommate A chooses rock 36% of the time, and roommate B chooses rock 22% of the time; roommate A selects paper 32% of the time, and roommate B selects paper 25% of the time; roommate A chooses scissors 32% of the time, and roommate B chooses scissors 53% of the time.

Let the probability that roommate A chooses rock = P(R_A) = 0.36

The probability that roommate A chooses paper = P(P_A) = 0.32

The probability that roommate A chooses scissors = P(S_A) = 0.32

The probability that roommate B chooses rock = P(R_B) = 0.22

The probability that roommate B chooses paper = P(P_B) = 0.25

The probability that roommate B chooses scissors = P(S_B) = 0.53

(a) Let A = event that roommate A wins the game and thus does not have to wash the dishes.

This will happen only when roommate A chooses rock and roommate B chooses scissors or roommate A chooses paper and roommate B chooses rock or roommate A chooses scissors and roommate B chooses paper.

So, P(A) =  P(R_A) \times P(S_B) + P(P_A) \times P(R_B) + P(S_A) \times P(P_B)  

             =  (0.36 \times 0.53) + (0.32 \times 0.22) + (0.32 \times 0.25)  

             =  0.1908 + 0.0704 + 0.08

    P(A)  =  0.34

(b) Let B = event that roommate B wins the game and thus does not have to wash the dishes.

This will happen only when roommate B chooses rock and roommate A chooses scissors or roommate B chooses paper and roommate A chooses rock or roommate B chooses scissors and roommate A chooses paper.

So, P(B) =  P(R_B) \times P(S_A) + P(P_B) \times P(R_A) + P(S_B) \times P(P_A)  

             =  (0.22 \times 0.32) + (0.25 \times 0.36) + (0.53 \times 0.32)  

             =  0.0704 + 0.09 + 0.1696

    P(B)  =  0.33

(c) Let C = event that the game ends in a tie.

This will happen only when roommate A chooses rock and roommate B also chooses rock or roommate A chooses paper and roommate B also chooses paper or roommate A chooses scissors and roommate B also chooses scissors.

So, P(C) =  P(R_A) \times P(R_B) + P(P_A) \times P(P_B) + P(S_A) \times P(S_B)  

             =  (0.36 \times 0.22) + (0.32 \times 0.25) + (0.32 \times 0.53)  

             =  0.0792 + 0.08 + 0.1696

    P(C)  =  0.3288 ≈ 0.33

(d) The complement of event A = P(A') = 1 - P(A)

                                                       = 1 - 0.34 = 0.66.

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