Assuming that s = 18,000 is a reasonable estimate for what sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years experience within $4200 if we wish to be 95% confident?

The minimum sample size required is 71.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean is:

The margin of error for this interval is:

It is provided that the population standard deviation σ can be estimated by the sample standard deviation s.

The given information is:

s = 18000,

Margin of error = $4200.

Confidence level = 95%

The z-value for 95% confidence level is,

*Use a z-table for the critical value.

Compute the sample size required as follows:

Thus, the minimum sample size required is 71.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b