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Goshia [24]
3 years ago
5

Susan and ronald dugan agreed upon the price of 256,000 for their new home. They plan to make 30 percent down payment and financ

e the rest at 7.5 persent for 20 years. What the total amount to be paid rounded to the nearest whole dollar?
Mathematics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

The total amount to be paid for new home is $838,014.72  

Step-by-step explanation:

Given as :

The price of the new house = $256,000

The down payment amount = 30% of house price

So, The down payment price =  30% of $256,000

i.e The down payment price =  \dfrac{30}{100} × 256000

Or, The down payment price = $76,800

Now, rest amount is finance

So, The finance Amount = p = $256000 - $76800 = $179,200

The rate of interest applied = r = 7.5%

The time period of finance = t = 20 years

Let The Amount after 20 years of finance = $A

Let The total amount to be paid for new home = $B

<u>Now, From Compound Interest </u>

Amount = Principal × (1+\dfrac{\textrm rate}{100})^{\textrm time}

Or, A = p × (1+\dfrac{\textrm r}{100})^{\textrm t}

Or, A = $179,200 × (1+\dfrac{\textrm 7.5}{100})^{\textrm 20}

Or, A = $179,200 × (1.075)^{\textrm 20}

Or, A = $179,200 × 4.24785

∴  A = $761,214.72

So,The Amount after 20 years of finance = A = $761,214.72

<u>Now, Again</u>

The total amount to be paid for new home = Down payment amount + The Amount after 20 years of finance

Or, B = $76,800 + A

Or, B = $76,800 + $761,214.72

Or, B = $838,014.72

So, The total amount to be paid for new home = B = $838,014.72

Hence, The total amount to be paid for new home is $838,014.72  Answer

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Answer:

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

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So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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