Use scenario 6-5.consider the following game. you pay me an entry fee of x dollars; then i roll a fair die. if the die shows a number less than 3 i pay you nothing; if the die shows a 3 or 4, i give you back your entry fee of x dollars; if the die shows a 5, i will pay you $1; and if the die shows a 6, i pay you $3. what value of x makes the game fair (in terms of expected value) for both of us?
a.$2
b.$4
c.$1
d.$0.75
e.$0.5

C.$1

Step-by-step explanation:

Since, when a die is rolled,

Then the total possible outcomes = 6 ( i.e. 1, 2, 3, 4, 5 or 6 )

Outcomes of getting number less than 3 = 2 ( i.e. 1 or 2 )

Outcomes of getting  3 or 4 = 2

Outcomes of getting 5 = 1,

Outcomes of getting 6 = 1,

Thus, the probability of getting number less than 3 =

The probability of getting 3 or 4 =

The probability of getting 5 =

The probability of getting 6 =

∵ If the die shows a number less than 3 then nothing will get; if the die shows a 3 or 4, we will get x dollars; if the die shows a 5, we will get $1; and if the die shows a 6, we will get $3

So, the expected value =

The game is fair if,

Hence, the value of x would be $ 1.

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

link is