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svetoff [14.1K]
1 year ago
12

The Office of Student Services at a large western state university maintains information on the study habits of its full-time st

udents. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a standard deviation of six hours. Suppose we select a random sample of 144 current students. What is the standard error of the mean? Select one: a. 0.50 b. 6.00 c. 0.25 d. 2.00
Mathematics
1 answer:
vredina [299]1 year ago
4 0

Answer:

a. 0.50

Step-by-step explanation:

The standard error of the mean is the standard deviation of the population divided by the square root of the sample size.

In this problem, we have that:

Standard deviation of the population: 6 hours

Sample size: 144

Square root of 144 is 12.

So the standard error of the sample mean is 6/12 = 0.5.

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A bucket that has a mass of 30 kg when filled with sand needs to be lifted to the top of a 30 meter tall building. You have a ro
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Step-by-step explanation:

We are given;

Mass of bucket = 30 kg

Mass of rope = 0.3 kg/m

height of building= 30 meter

Now,

work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand

Or W = W1 + W2

Work done in lifting the rope is given as,

W1 = Force x displacement

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Integrating, we have;

W1 = [0.2x²/2] at boundary of 30 and 0

W1 = 0.1(30²)

W1 = 90 J

work done in lifting the sand is given as;

W2 = (30,0)∫(F .dx)

F = mx + c

Where, c = 30 - 15 = 15

m = (30 - 15)/(30 - 0)

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So,

F = 0.5x + 15

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W2 = (30,0)∫(0.5x + 15 .dx)

Integrating, we have;

W2 = (0.5x²/2) + 15x at boundary of 30 and 0

So,

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Therefore,

work done lifting the bucket (sand and rope) to the top of the building,

W = 90 + 675

W = 765 J

5 0
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