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

Simplify

Evaluate

Hence, the expected value of the game is $1
Read more about expected values at:
brainly.com/question/15858152
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.
Its simple.
Cost of t-Shirt each=$14
Cost of shorts each=$18
No. of clothes=x t-shirts and y shorts
so just multiply cost of each t-shirt no.of t-shirts and do the same thing with the Shorts
you get the answer as 14x +18y
There! you've expressed it in the form of ax+by and 14 and 18 are also positive or negative constant
Youve found the answer satisfing all the requirements
hope it helped you aand if it did please mark as brainliest answer
Thanks and keep studying
The correct answer is ∠Y = 75°
because
45°∠A ∼ ∠X
∠B ∼ ∠Y = 75°
60°∠C ∼ ∠Z
45 + 60 = 105
180 -105 = 75
I believe it is 2/12, because they each have a 1/6 probability of rolling a 6.