Here is the work for that problem
A rectangular prism has a length of 9 inches, a height of 5 inches, and a width of 7
1 answer:
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9514 1404 393
Answer:
- $33,336.83
- $35,176.68
- $36,869.23
- $38,266.75
- $39,158.08
Step-by-step explanation:
The amortization formula can be used to find the payment value.
A = P(r/2)/(1 -(1 +r/2)^(2n))
where P is the principal amount at the beginning of the loan period, r is the annual interest rate, n is the number of years remaining on the loan
The value of A in this scenario is the semi-annual payment.
Initially, the interest rate is 0.055 and the number of years remaining is 25.
After the first 5-year period, the interest rate goes up by 12%, so is multiplied by a factor of 1.12 to become 6.16% per year. The same growth factor is applied to the interest rate at the beginning of each 5-year period.
Of course, the number of years remaining is decreased by 5 years at the beginning of the next 5-year period.
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The Principal remaining at the end of each 5-year period is the starting principal for the next period. It is calculated from ...
FV = P(1 +r/2)^10 -A((1 +r/2)^10 -1)/(r/2)
Where P is the starting principal, A is the loan payment as calculated above, and r is the annual interest rate.
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These formulas are built into spreadsheet functions, so the desired set of loan payments can be calculated easily by that technology. The result is attached. In the "# pmts" column is the value used to amortize the loan for the 5-year period. The semiannual payment is calculated as though the loan would be completely paid off in that number of payments, keeping the same interest rate for the duration. Of course, that payment series is interrupted and the loan recalculated at the beginning of the next 5 years.
The half-yearly payments for each 5-year interval are ...
$33,336.83
$35,176.68
$36,869.23
$38,266.75
$39,158.08
_____
<em>Additional comment</em>
The values displayed in the spreadsheet are rounded to the values shown. The values used in calculation have 14 or more significant digits. This means the numbers here may vary from those provided by a lending institution. In the real world, the principal and interest values are rounded to cents with each payment, so actual results may vary by a few cents either way.
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.