Find the probability a random voltage is between 124.1 volts and 124.7 volts.

A. 0.55

B. 0.15

C. 0.30

D. 0.45

C. 0.30

The probability is 0.30

Step-by-step explanation:

The diagram used in solving this question is attached.

From the given graph that is attached: the needed region have

height = 0.5

width = 124.7 - 124.1 = 0.6

The area of the rectangular region is:

Therefore, the probability of a random voltage between 124.1 and 124.7 is 0.3

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.