The probability that mary will win a game is 0.02, so the probability that she will not win is 0.98. if mary wins, she will be given $160; if she loses, she must pay $16. if x = amount of money mary wins (or loses), what is the expected value of x? (round your answer to the nearest cent.)

Given:

Probability of winning, P(X) = 0.02

Probability of losing, P() = 0.98

Wining amount = $160

Losing amount = $16

Step-by-step explanation:

Let the expected amount of money win be  'X'

Expected value of X, E(X) = Probability of winning, P(X).Probability of winning, P(X)  - Probability of losing, P().Losing amount

Now,

E(X) = ()

E(X) = -12.48

Expected value of X = -12.48

Expected loss value = $12.48 loss

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.