The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the
management is that after this promotional policy is advertised, the expenditures for all customers at this
supermarket will be normally distributed with a mean of $95 and a standard deviation of $16. If the
management wants to give free gifts to at most 13.57% of the customers, what should the amount be
above which a customer would receive a free gift?

X ≥ 112,6 $

Step-by-step explanation:

Normal Distribution  N ( 95 , 16 )

μ₀ = 95

σ  =  16

We need to get z (score) for area 0,1357; then from z - Table we get

z (score) = 1,1

Now  z (score ) = ( X - μ₀ )/σ

1,1 = ( X - 95 ) / 16

17,60 + 95  = X

X = 112,6 $

That means that in order to get a gift customes has to spend at least 112,6 $

if she was to only spend a dollar she can get 1lb

step-by-step explanation:

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.