You still work at Dockery Manufacturing (still exciting!), and your organization still requires a lead time between 18 and 23 days for supplied items. After further investigating the supplier’s process, you determine their current lead times are normally distributed with a process mean of 21 days and a standard deviation of 1.90. What is the percent likelihood that a randomly observed lead time by the supplier would be too short per your firm’s specifications? Group of answer choices 5.71% 14.69% 20.58% 35.13% 44.29%

The probability is of 14.69%

Step-by-step explanation:

According to the given data we have the following:

Lead time mean = 21 days

Standard deviation = 1.90

To calculate the percent likelihood that a randomly observed lead time by the supplier would be too short per your firm’s specifications we will calculate first the z statisticsas follows:

z= (X- mean)/ Standard deviation

z = (23-21)/ 1.90

z = 1.05

Then Looking at z table, we will find probability corresponding to z = 1.05. The probability is 0.8531 but the area for z > 1.05 will be 1- 0.8531= 0.1469

Therefore, the probability is of 14.69%

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.

about 168 miles

step-by-step explanation: