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3 years ago

A pyramid has a square base with sides of length s. The height of the pyramid is eqaul to 1/2 of the length of a side base. Whic

h formula represents the volume of a pyramid?
A) V=1/12s^2
B) V=1/6s^3
C) V=1/3s^3
D) V=3s^3
E) V=6s^3

Mathematics

1 answer:

Paha777 [63]3 years ago

8 0

$A\ volume\ of\ a\ square\ pyramid:\\\\V=\dfrac{1}{3}a^2H\\\\a-length\ of\ a\ side\ of\ a\ base\\H-a\ height\ of\ a\ pyramid\\-------------------\\a=s;\ H=\dfrac{1}{2}s\\\\subtitute\\\\V=\dfrac{1}{3}s^2\cdot\dfrac{1}{2}s=\dfrac{1}{6}s^3\\\\\boxed{V=\frac{1}{6}s^3}\to\fbox{B.}$

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Step-by-step explanation:

There are around 52 weeks in a year. You make $348 a week.

52 x 348 = $18,096

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Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

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Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago

A regression analysis between weight (y in pounds) and height (x in inches) resulted in the following least squares line: ŷ = 13

Lunna [17]

Answer:

$y=6x +135$

And for this case the interpretation for the slope would be that for every unit that the height in inches increase then the weight in pounds increase 6 units.

For the intercept of 135 represent the amount initial amount of weight for the scale.

Step-by-step explanation:

We assume that they use least squares in order to create the regression equation

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

And the slope would be:

$m=6$

The means for x and y are given by:

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

$\bar y= \frac{\sum y_i}{n}$

And we can find the intercept using this:

$b=\bar y -m \bar x$

For this case we know that the line adjusted is:

$y=6x +135$

And for this case the interpretation for the slope would be that for every unit that the height in inches increase then the weight in pounds increase 6 units.

For the intercept of 135 represent the amount initial amount of weight for the scale.

3 0

3 years ago