A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Suppose that 90% of all batteries from a certain supplier have acceptable voltages. Among twenty randomly selected flashlights, what is the probability that at least nineteen will work? What assumptions did you make in the course of answering the question posed? a. we have assumed here that the batteries' voltage levels are independent. b. we have assumed here that the batteries' voltage levels are dependent.

a) 0.064

b) 0.002

Step-by-step explanation:

a) Assuming that  the batteries' voltage levels are independent

Probability = 0.8 * 0.8 = 0.64

b) batteries' voltage levels are dependent

20/19 (0.064)^19 *0.036  + 20/20 (0.064)^20 *0.036^0

= 0.002

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah

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.