MATH SOLVE

2 months ago

Q:
# A house was valued at $120,000 in the year 1992. The value appreciated to $160,000 by the year 2007.Use the compund interest formula S=P(1+r)t to answer the following questions.A) What was the annual growth rate between 1992 and 2007?r = ______ Round the growth rate to 4 decimal places.B) What is the correct answer to part A written in percentage form?r =___ %.C) Assume that the house value continues to grow by the same percentage. What will the value equal in the year 2010 ?value = $ ____ Round to the nearest thousand dollars

Accepted Solution

A:

Answer:A) 0.0194B) 1.94%C) $ 170,000Step-by-step explanation:Value of house in 1992 = P = $ 120,000Value of house in 2007 = S = $ 160,000Time difference from 1992 to 2007 = 15 yearsPart A)The formula of compound interest is:[tex]S=P(1+r)^{t}[/tex]P is the original amount i.e. $ 120,000t is the time in years which is 15 yearsS is the amount after t years which is $ 160,000r is the annual growth rateUsing the values, we get:[tex]160000=120000(1+r)^{15}\\\\\frac{160000}{120000}=(1+r)^{15}\\\\ \frac{4}{3}=(1+r)^{15}\\\\(\frac{4}{3} )^{\frac{1}{15}}=1+r\\\\ r=(\frac{4}{3} )^{\frac{1}{15}}-1\\\\ r=0.0194[/tex]Thus, the annual growth rate is 0.0194Part B)In order to convert a decimal to percentage, simply multiply the decimal by 100.So, 0.0194 in percentage would be 1.94%Part C)We have to find the value of house in 2010 i.e. after 18 years. So t =18Using the values in the formula, we get:[tex]S=120000(1+0.0194)^{18}\\\\ S=169584[/tex]Rounded to nearest thousand dollars, the value of the house would be $ 170,000