The following table shows the probability distribution for the prize amounts that will be awarded at a school raffle. Prize $1 $5 $10 $20 $50 Probability 0.60 0.30 0.05 0.04 0.01 Let the random variable P represent a randomly selected prize amount. What is the expected value of P ? $1.00 A $3.90 B $4.00 C $10.00 D $17.20

This question is based on the concept of probability. Therefore, the correct option is A , $3.90 is the expected value of P.

Given:

Prize                 $1          $5            $10            $20              $50

Probability     0.60      0.30         0.05           0.04              0.01

We need to determined the the expected value of P.

According to question,

By using the definition of expected value of probability.

The expected value is defined as the the sum of each outcome is multiplied by its probability.

Therefore, the formula of expected value,

Therefore, the correct option is A, i.e. $ 3.90 is the expected value of P.

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$3.90

Step-by-step explanation:

The expected value is the sum of each outcome multiplied by its probability.

E(P) = (1)(0.60) + (5)(0.30) + (10)(0.05) + (20)(0.04) + (50)(0.01)

E(P) = 3.9

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer

a) at most means it can cost $12 or less

so the inequality sign would be ≤

you would then need to multiply the number of toppings by the 0.35 for each one and add it to the cost of the cheese pizza.

the equation would be 9.99 + 0.35x ≤ 12

b) using the above equation:

9.99 + 0.35x ≤ 12

subtract 9.99 from each side:

0.35x ≤2.01

divide both sides by 0.35:

x ≤ 2.01 / 0.35

x ≤5.74

at most you could have 5 toppings.

the 5 topping pizza would cost: 0.35 x 5 = 1.75 + 9.99 = $11.74