Mr. adams and ms. smith are betting on repeated flips of a coin. at the start of the game mr. adams has a dollars and ms. smith has b dollars, at each flip the loser pays the winner one dollar, and the game continues until either player is “ruined.” making use of the fact that in an equitable game each player’s mathematical expectation is zero, find the probability that mr. adams will win ms. smith’s b dollars before he loses his a dollars.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4