Respuesta :
Givn:
Value of the car in 1995 = $32,000
Value of the car in 2001 = $14,000
Let's solve for the following:
• (A). What was the annual rate of change between 1995 and 2001?
Apply the exponential decay formula:
[tex]f(t)=a(1-r)^t[/tex]Where:
• t is the number of years between 2001 and 1995 = 2001 - 1995 = 6
,• a is the initial value = $32000
,• r is the rate of decay.
,• f(t) is the present value
Thus, we have
[tex]\begin{gathered} 14000=32000(1-r)^6 \\ \end{gathered}[/tex]Divide both sides by 32000:
[tex]\begin{gathered} \frac{14000}{32000}=\frac{32000(1-r)^6}{32000} \\ \\ 0.4375=(1-r)^6 \end{gathered}[/tex]Take the 6th root of both sides:
[tex]\begin{gathered} \sqrt[6]{0.4375}=\sqrt[6]{(1-r)^6} \\ \\ 0.87129=1-r \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} r=1-0.87129 \\ \\ r=0.1287 \\ \\ r=0.1287*100=12.87\text{ \%} \end{gathered}[/tex]Therefore, the rate of change between 1995 and 2001 is 0.1287
• (B). What is the correct answer to part A written in percentage form?
In percentage form, the rate of change is 12.87 %
• (C),. Assume that the car value continues to drop by the same percentage. What will the value be in the year 2005?
We have the equation which represents this situation below:
[tex]f(t)=32000(1-0.1287)^t[/tex]Here, the value of t will be the number of years between 1995 and 2005.
t = 2005 - 1995 = 10
Now, substitute 10 for t and solve for f(10):
[tex]\begin{gathered} f(10)=32000(1-0.1287)^{10} \\ \\ f(10)=32000(0.8713)^{10} \\ \\ f(10)=32000(0.25216) \\ \\ f(10)=8069.14\approx8100 \end{gathered}[/tex]Therefore, the value in the year 2005 rounded to the nearest 50 dollars is $8100
ANSWER:
• (a). 0.1287
,• (b). 12.87%
,• (c). $8100
