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Deffense [45]
3 years ago
14

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between 2000

0 and 35000 . Assume that a confidence interval estimate of the population mean annual starting salary is desired.
What is the planning value for the population standard deviation?
Mathematics
1 answer:
ehidna [41]3 years ago
7 0

Answer:

The planning value for the population standard deviation is of 4330.

Step-by-step explanation:

Uniform probability distribution:

The uniform probability distribution has two bounds, a and b. The standard deviation is given by:

S = \sqrt{\frac{(b-a)^2}{12}}

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between 20000 and 35000.

Uniform in this interval, so a = 20000, b = 35000

What is the planning value for the population standard deviation?

S = \sqrt{\frac{(35000 - 20000)^2}{12}} = 4330

The planning value for the population standard deviation is of 4330.

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X+2y≤6  with x≥0 and y≥0 
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1st) Find the 1st  shaded region: 1st Quadrant since x≥0 and y≥0

2nd) Let's find the x and y intercepts of 2x+y = 6
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3rd) Let's find the x and y intercepts of x+2y = 6
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Re write all pairs:

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For A(3,6), →→F=15+12 = 25
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Hope that I didn't make any mistake

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Hope this helps!
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