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DochEvi [55]
2 years ago
13

Can someone help me with answer C? its the last one i need

Mathematics
1 answer:
Dmitry [639]2 years ago
4 0

Using the monthly payment formula, it is found that her down payment should be of $1,419.

<h3>What is the monthly payment formula?</h3>

It is given by:

A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}

In which:

  • P is the initial amount.
  • r is the interest rate.
  • n is the number of payments.

For this problem, the parameters are:

A = 250, r = 0.072, n = 72.

Hence:

r/12 = 0.072/12 = 0.006.

We solve for P to find the total amount of the monthly payments, hence:

A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}

P\frac{0.006(1.006)^{72}}{(1.006)^{72}-1} = 250

0.0171452057P = 250

P = 250/0.0171452057

P = $14,581.

The total payment is of $16,000, hence her down payment should be of:

16000 - 14581 = $1,419.

More can be learned about the monthly payment formula at brainly.com/question/26476748

#SPJ1

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A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount
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Answer:

x = 97

Step-by-step explanation:

Given

t = 20 --- time (years)

A =1000 --- amount

r = 10\% --- rate of interest

Required

The last 10 payments (x)

First, calculate the end of year 1 payment

y_1(end) = 10\% * 1000 * 150\%

y_1(end) = 150

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A_1=A - y_1(end) - r * A

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Rewrite as:

A_2 = 0.95 * 1000^2

We have been able to create a pattern:

A_1 = 1000 * 0.95^1 = 950

A_2 = 1000 * 0.95^2 = 902.5

So, the payment till the end of the 10th year is:

A_{10} = 1000*0.95^{10}

A_{10} = 598.74

To calculate X (the last 10 payments), we make use of the following geometric series:

Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n

Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n

Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n

Amount = \sum\limits^{9}_{n=0} x * (1.10)^n

The amount to be paid is:

Amount = A_{10}*(1 + r)^{10} --- i.e. amount at the end of the 10th year * rate of 10 years

Amount = 1000 * 0.95^{10} * (1+r)^{10}

So, we have:

Amount = \sum\limits^{9}_{n=0} x * (1.10)^n

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}

The geometric sum can be rewritten using the following formula:

S_n = \sum\limits^{9}_{n=0} x * (1.10)^n

S_n =\frac{a(r^n - 1)}{r -1}

In this case:

a = x

r = 1.10

n =10

So, we have:

\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n

\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n

\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n

x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n

So, the equation becomes:

x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}

Solve for x

x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}

x = 97.44

Approximate

x = 97

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