A house was valued at $100,000 in the year 1987. The value appreciated to $165,000 by the year 2002.

Use the compound interest formula S=P(1+r)^t to answer the following questions.
A) What was the annual growth rate between 1987 and 2002? (Round to 4 decimal places.)
B) What is the correct answer to part A written in percentage form?
C) Assume that the house value continues to grow by the same percentage. What will the value equal in the year 2005? (Round to the nearest thousand dollars.)

To answer these questions, we will use the compound interest formula S = P(1 + r)^t, where:

- S is the final value of the house
- P is the initial value of the house
- r is the annual growth rate expressed as a decimal
- t is the number of years

A) To find the annual growth rate between 1987 and 2002, we have the following values:
- P = $100,000 (initial value in 1987)
- S = $165,000 (final value in 2002)
- t = 2002 - 1987 = 15 years

We can rearrange the formula to solve for r:
r = ((S / P)^(1/t)) - 1

Plugging in the values, we have:
r = (($165,000 / $100,000)^(1/15)) - 1

Calculating this expression, we find that the annual growth rate (r) between 1987 and 2002 is approximately 0.0417.

B) To express the annual growth rate in percentage form, we multiply it by 100 and round to 4 decimal places:
Percentage form = 0.0417 * 100 = 4.17%.

Therefore, the correct answer to part A written in percentage form is 4.17%.

C) To find the value of the house in the year 2005 assuming the same growth rate, we have:
- P = $165,000 (final value in 2002)
- t = 2005 - 2002 = 3 years

Using the formula S = P(1 + r)^t and plugging in the values, we get:
S = $165,000 * (1 + 0.0417)^3

Calculating this expression, we find that the value of the house in the year 2005 would be approximately $190,912 (rounded to the nearest thousand dollars).

Therefore, the value of the house in the year 2005 would be $191,000.

100,000(1+r)^15 = 165,000

(1+r)^15 = 1.65
take the 15th root of both sides using your calculator.

let me know what you get.