A traffic light at a certain intersection is green 50% of the time, yellow 10% of the time, and red 40% of the time. A car approaches this intersection once each day. Let X represent the number of days that pass up to and including the first time the car encounters a red light. Assume that each day represents an independent trial. Find P(X ≤ 3).

step-by-step explanation:

4227473658779875563886538705958641728563470543621287463824132573847969p05749365273847906970680574936214

eve and marks ratios are equivalent

step-by-step explanation:

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g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.