The alphabet game costs $.25 to play. Before the game, 26 slips of paper with a different letter of the alphabet on it are put into a bag. A player draws one slip from the bag. If the player draws a vowel (A, E, I, O, or U), he or she wins $1. If a player plays the alphabet game 2 times in a row, replacing the slip of paper after each turn, what is the probability that they win twice? Write as a fraction.

25/676

Step-by-step explanation:

Number of possible outcomes = 26

In other to win, one must draw must be either (A, E, I, O or U)

Therefore required drws to win = 5

First draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Second draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Therefore,

P(winning twice) = (5/26) × (5/26) = 25/676

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

idk

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