Thickness measurements of a coating process are made to the nearest hundredth of amillimeter. The thickness measurements are uniformly distributed with values 0.15, 0.16, 0.17, 0.18, and 0.19. Determine the mean and variance of the coating thickness for this process. A. mean (in millimeter).
B. variance (in millimeters^2).

Mean=

Variance=

Step-by-step explanation:

Given is the uniform distribution

x=0.15,y=0.19

where x and y are lower and upper values.

mean=(x+y)/2

mean=(0.19+0.15)/2

Mean=0.17 mm

Now using equation

variance =

For variance take a=15,b=19

Variance=

Variance=2

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.