Find the length of a line segment with endpoints of -9 and 7.
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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
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.
9514 1404 393
Answer:
252.8 cm²
Step-by-step explanation:
The missing side of the right triangle can be found from the Pythagorean theorem:
s² = 20² -16² = 400 -256 = 144
s = 12 . . . . cm
The area of a right triangle is more easily found using the traditional area formula:
A = 1/2bh
A = 1/2(12 cm)(16 cm) = 96 cm² (left-side triangle)
The area of the triangle on the right can be found from Heron's formula. The semiperimeter is ...
s = (16 +20 +23)/2 = 29.5
The area is ...
A = √(29.5(29.5 -16)(29.5 -20)(29.5 -23)) = √(29.5·13.5·9.5·6.5)
A = √24591.9375 ≈ 156.818 . . . . . cm² (right-side triangle)
Then the total area of the figure is ...
A = 96 cm² +156.818 cm² = 252.818 cm² . . . . total area
