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Ilya [14]
3 years ago
5

Daphne bought a house for 335,000. She financed 276,475 of the purchase price with a 30 year, fixed rate mortgage with 5.65% int

erest rate. What is the total cost of the principal and interest after 30 years
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
8 0
First find the yearly payment using the formula of the present value of annuity ordinary
The formula is
Pv=pmt [(1-(1+r)^(-n))÷r]
Pv present value 276475
Pmt yearly payment ?
R interest rate 0.0565
N time 30 years

Now solve for pmt
The formula change to be
Pmt=pv÷ [(1-(1+r)^(-n))÷r]
Plug in the equation above
Pmt=276,475÷((1−(1+0.0565)^(−30))÷(0.0565))=19,339.22

Now find the cost of the principle and interest after 30 years by multiplying the yearly payment by the time

19,339.22×30=580,176.60...answer

Hope it helps:-)
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A particular psychological test is used to measure need for achievement. The average test score for all university students in O
Angelina_Jolie [31]

Answer:

Only B and C are always true.

Step-by-step explanation:

To analyse which statements are always true, we go through the process of finding confidence intervals for sample means, from the start.

Confidence Interval = (Sample mean) ± (Margin of error)

From this expression, it is evident that the margin of error determines how wide the confidence interval would be.

Sample Mean = 110 (given)

Margin of Error = (Critical value) × (Standard deviation of the distribution of sample means)

Since no information about the population standard deviation is provided, the critical value is obtained using t-distribution.

The critical value usually varies at different confidence levels and degree of freedoms.

The higher the confidence level, the higher the critical value and the higher the margin of error leading to a wider range.

Hence, a confidence interval of 95% will have a higher critical value than a confidence interval of 90%. Hence, statement C is proved once that 'for n = 100, the 95% confidence interval will be wider than the 90% confidence interval'.

After obtaining the critical value, we then obtain the standard deviation of the distribution of sample means or simply the standard error of the mean. This is given as

σₓ = σ/√n

where σ = standard deviation; which isn't given. The standard deviation might be high enough to guarantee that the Margin of error is high too for the confidence interval to contain 115 or low enough to ensure that the Margin of error is very small and the confidence interval will not contain 115.

Or the sample size might be high enough to make the standard error of the mean to be eventually small and lead to a small margin of error and the condidence interval will not contain 115.

The point is, it isn't always sure that the resulting interval.would contain 115. So, statement A isn't always true.

Then from σₓ = σ/√n,

n = sample size, a large sample size means a more narrow confidence interval and a small sample size means a wider sample size. This proves statement C.

The 95% confidence interval for n = 100 will be more narrow than the 95% confidence interval for n = 50.

Hence, Only B and C are always true.

Hope this Helps!!!

5 0
3 years ago
If someone can help me out :D!
Nataliya [291]
Hello

Answer:

28.378

Explanation:

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The model shown represents which of the following?
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3 years ago
The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 3639 3639 miles, with a var
Andrei [34K]

Answer:

96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

A reminder is that the standard deviation is the square root of the variance.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

\mu = 3639, \sigma = \sqrt{145161} = 381, n = 41, s = \frac{381}{\sqrt{41}} = 59.5

Probability that the mean of the sample would differ from the population mean by less than 126 miles

This is the pvalue of Z when X = 3639 + 126 = 3765 subtracted by the pvalue of Z when X = 3639 - 126 = 3513. So

X = 3765

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{3765 - 3639}{59.5}

Z = 2.12

Z = 2.12 has a pvalue of 0.983

X = 3513

Z = \frac{X - \mu}{s}

Z = \frac{3513 - 3639}{59.5}

Z = -2.12

Z = -2.12 has a pvalue of 0.017

0.983 - 0.017 = 0.966

96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles

3 0
3 years ago
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