MATH SOLVE

2 months ago

Q:
# A car was valued at $39,000 in the year 1995. The value depreciated to $11,000 by the year 2003.A)What was the annual rate of change between 1995 and 2003? (Round to 4 decimal places)B)What is the correct answer to part A written in percentage form?C)Assume that the car value continues to drop by the same percentage. What will the value be in the year 2007?Please help ASAP the homework is due Monday!!! :(

Accepted Solution

A:

Answer:14.6328% , $5836.03 Step-by-step explanation:Here we are going to use the formula [tex]A_{0}(1-r)^n = A_{n}[/tex][tex]A_{0}[/tex] = 39000r=?[tex]A_{8}[/tex] = 11000n=8Hence [tex]39000(1-r)^8 = 11000[/tex][tex](1-r)^8 = \frac{11000}{39000}[/tex][tex](1-r)^8 = 0.2820[/tex][tex](1-r) = 0.2820^{\frac{1}{8}[/tex][tex](1-r) = 0.2820^{0.125}[/tex][tex](1-r) = 0.8536[/tex][tex](1-0.8536=r[/tex][tex]r = 0.1463[/tex]Hence r= 0.1463In percentage form r = 14.63%Now let us see calculate the value of car in 2003 that is after 12 yearswe use the main formula again [tex]A_{0}(1-r)^n = A_{n}[/tex][tex]A_{0}[/tex] = 39000r=0.1463[tex]A_{12}[/tex] = ?n=12[tex]39000(1-0.14634)^{(12} = A_{12}[/tex][tex]39000(0.8536)^{12} = A_{12}[/tex][tex]39000*0.1497 = A_{12}[/tex][tex]A_{12}=5840.34[/tex]Hence the car's value will be depreciated to $5840.34 (approx) by 2003.