The lengths of plate glass parts are measured to the nearest tenth of a millimeter. the lengths are uniformly distributed with values at every tenth of a millimeter starting at 590.2, and continuing through 590.9. determine the mean and variance of the lengths. (a) mean (in tenths of millimeters) round your answer to two decimal places (e. g. 98.76). (b) variance (in tenths of millimeters2) round your answer to three decimal places (e. g. 98.765).

Step-by-step explanation:

Given : The lengths are uniformly distributed with values at every tenth of a millimeter starting at 590.2, and continuing through 590.9.

Let x be the random variable that represents the lengths of plate glass parts.

The mean and variance of the random variable X that is uniformly distributed in interval [a,b] is given by :-

In the given situation, a= 590.2 and b= 590.9

Then,

Hence, Mean = 590.55 and Variance = 0.41

we need the options

step-by-step explanation:

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi