3 = –(–y + 6) a. –9 b. –3 c. 3 d. 9

1 answer:

7 0

3 = -(-y + 6)

First, simplify brackets. / Your problem should look like: 3 = y - 6
Second, add 6 to both sides. / Your problem should look like: 3 + 6 = y
Third, simplify 3 + 6 to 9. / Your problem should look like: 9 = y
Fourth, switch sides. / Your problem should look like: y = 9

The answer is D) 9.

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Consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number,

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The expected value of the points earned on a single roll in this game is $\dfrac{1}{6} = 0.1667$ .

Step-by-step explanation:

We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.

Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).

Here, the prime numbers are = 1, 2, 3 and 5

Numbers which are not prime = 4 and 6

This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.

Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.

Note that the probability of getting any of the specific number on the dice is $\dfrac{1}{6}$ .

Numbers on the dice (X) P(X)

+1 $\frac{1}{6}$

+2 $\frac{1}{6}$

+3 $\frac{1}{6}$

-4 $\frac{1}{6}$

+5 $\frac{1}{6}$

-6 $\frac{1}{6}$

Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.

Now, the expected value of X, E(X) = $\sum X \times P(X)$

= $(+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}$