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A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)
Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.
The lateral faces are equilateral triangles of side length x.
The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.
now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.
by the Pythagorean theorem:
thus,
thus:
(cm^2)
but
is exactly the base area, since the base is a square of sidelength = x cm.
So, the total surface area = base area + lateral area =
cm^2
Answer: cm^2
(and the lateral edges of the pyramid are also equal to each other)
Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.
The lateral faces are equilateral triangles of side length x.
The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.
now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.
by the Pythagorean theorem:
thus,
thus:
but
So, the total surface area = base area + lateral area =
Answer:
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
Then, we need to subtract one by the confidence level and divide by 2. So:
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 .
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
Now, we multiply T and s
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
So
With 119 and 0.025 in the t-distribution table, we have .
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
So
With 119 and 0.025 in the t-distribution table, we have .
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.