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3 years ago

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A researcher wants to compare student loan debt for students who attend four-year public universities with those who attend four

–year private universities. She plans to take a random sample of 100 recent graduates of public universities and 100 recent graduates of private universities. Which type of random sampling is utilized in her study design?

1 answer:

dlinn [17]3 years ago

3 0

Answer:

A simple random sample.

Step-by-step explanation:

A simple random sample is an statistical sample in which each member of a group has the same probability of being chosen. Since the researcher doesn't really have specific characteristics added to the sample other than being from public or private universities, this would be a simple random sample.

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Step-by-step explanation:

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How do you find the volume of the solid generated by revolving the region bounded by the graphs

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Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.

We can find the area like this:

$A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy$

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= \pi \int_{0}^2 64-32x^2 +4x^4 dx$

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \frac{\pi}{2} \int_{0}^8 y dy$

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

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One of the events in the Winter Olympics is the Men's 500-meter Speed Skating.

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Answer:

Mean: $\bar x = 40.61$

$Median =40.2$

$Mode = 43.4$

Step-by-step explanation:

Given

See attachment for data

Solving (a): The mean

Mean is calculated as:

$\bar x = \frac{\sum x}{n}$

From the attached:

$n = 14$

So, the mean is:

$\bar x = \frac{43.4+43.4+43.4+43.1+43.2+40.2+40.2+40.1+40.3+39.44+39.17+38.03+38.19+36.45}{14}$

$\bar x = \frac{568.58}{14}$

$\bar x = 40.61$

Solving (b): The median

$n = 14$ ; this is an even number. So, the median is:

$Median = \frac{1}{2}(n+1)$

$Median = \frac{1}{2}(14+1)$

$Median = \frac{1}{2}(15)$

$Median = 7.5th$

This implies that the median is the average of the 7th and 8th item.

Next, is to order the data (in ascending order): <em>36.45, 38.03, 38.19, 39.17, 39.44, 40.1, 40.2, 40.2, 40.3, 43.1, 43.2, 43.4, 43.4, 43.4.</em>

The 7th and 8th items are: 40.2 and 40.2

The median is:

$Median = \frac{1}{2}(40.2 + 40.2)$

$Median = \frac{1}{2}*80.4$

$Median =40.2$

Solving (c): The mode

43.4 has the highest number of occurrence.

So:

$Mode = 43.4$

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3 years ago