The expected value of the game is the mean value of the game
The expected value of the game is $1
<h3>How to determine the expected value?</h3>
There are 13 spades in a deck of card of 52
So, the probability of selecting a spade is:
P(Spade) = 13/52
Simplify
P(Spade) = 1/4
Winning = $7
The probability of not selecting a spade is:
P(Not spade) = 1 - 1/4
Simplify
P(Not spade) = 3/4
Lose = $1
The expected value of the game is:

This gives
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Simplify

Evaluate

Hence, the expected value of the game is $1
Read more about expected values at:
brainly.com/question/15858152
The answer is B -4 so ya I think though
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In order to find the x-intercept we need to make y=0
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Then we isolate the x
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
x y Ordered Pair type of intercept
0 0 (0,0) x-intercept
Now for the y-intercept we need to make x=0

we isolate the y
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
0 0 (0,0) y-intercept
ANSWER
0 0 (0,0) x-intercept
Answer:D
Step-by-step explanation: