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3 years ago

12

Look at picture for the question.

1 answer:

s344n2d4d5 [400]3 years ago

8 0

The answer of the following question is 5.8

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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

Alja [10]

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

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

$y_1(end) = 150$

Amount at end of year 1

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

$A_1=1000 - (150 - 10\% * 1000)$

$A_1 =1000 - (150- 100)$

$A_1 =950$

Rewrite as:

$A_1 = 0.95 * 1000^1$

Next, calculate the end of year 1 payment

$y_2(end) = 10\% * 950 * 150\%$

$y_2(end) = 142.5$

Amount at end of year 2

$A_2=A_1 - (y_2(end) - r * A_1)$

$A_2=950 - (142.5 - 10\%*950)$

$A_2 = 902.5$

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

$x = 97$

4 0

3 years ago

Building a has 7,500 ft of an office space for 320 employees building b has 9,500 square feet of office space for 370 employees

iren2701 [21]

Answer:

I believe the correct answer is Building B

Step-by-step explanation:

Building A is 23.45 per employee and Building B is 25.67 per employee

4 0

3 years ago

You want to make a compost box bigger than the basic 1-2-3 foot size. Your friend says, "If you double each dimension, you'll be

lisov135 [29]

No it'll be way more. just do the math showing the capacity with the original and doubled dimensions

7 0

3 years ago

HELP! last one of the day!

quester [9]

The bisector of the angle at A (call it AQ) divides the segment BC into segments BQ:QC having the ratio AB:AC. Use this fact to find x.

.. 9:15 = (2x -1):3x
.. 15(2x -1) = 9*3x . . . . . the product of the means equals the product of extremes
.. 30x -15 = 27x
.. 3x = 15
.. x = 5

___
According to the value of x, the bisector AQ divides the triangle into two isosceles triangles: ABQ, ACQ.

7 0

4 years ago

Help this is due soon

vichka [17]

Step-by-step explanation:

22x13, 10x9, then both answers from the previous equations multiplied. Any questions?

8 0

3 years ago