This is a hw problem i have

Question

A house was valued at $115,000 in the year 1990. The value appreciated to $165,000 by the year 2005.Use the compund interest formula  to answer the following questions.A) What was the annual growth rate between 1990 and 2005? =  Round the growth rate 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 2008 ?value = $  Round to the nearest thousand dollars.

Answer 1

165/115 = (1+r)^15

(1+r)^15 = 33/23 = about 1.43

(1+.02)^15 = about 1.39

the interest rate is slightly over 2%

1+r = the 15th root of 33/23

r = (33/23)^(1/15) -1 = about .0243595 = about .0244 rounded to 4 decimals

= 2.44% rate of growth annually

in 3 more years

x =115000 (1.0243595)^18

x = $177,354.07 in 2008

= about $177,000