Question

homework assignment 6.1 exponential functions
due sunday by 11:59pm points 12 submitting an external tool
homework assignment 6.1 exponential functions
score: 6/12 answered: 6/12
question 7
a car was valued at $32,000 in the year 1991. the value depreciated to $14,000 by the year 2001.
a) what was the annual rate of change between 1991 and 2001?
r = round the rate of decrease to 4 decimal places.
b) what is the correct answer to part a written in percentage form?
r = %.
c) assume that the car value continues to drop by the same percentage. what will the value be in the year 2004?
value = $ round to the nearest 50 dollars.
question help: video read written example

Explanation:

Step1: Use the exponential decay formula

The exponential decay formula is \(y = a(1 - r)^t\), where \(a=\$32000\), \(y = \$14000\), and \(t=2001 - 1991=10\) years.
Substitute the values into the formula: \(14000=32000(1 - r)^{10}\).

Step2: Solve for \(r\)

First, divide both sides by \(32000\): \(\frac{14000}{32000}=(1 - r)^{10}\).
Simplify \(\frac{14000}{32000}=\frac{7}{16}\). So, \(\frac{7}{16}=(1 - r)^{10}\).
Take the \(10\) - th root of both sides: \((\frac{7}{16})^{\frac{1}{10}}=1 - r\).
Calculate \((\frac{7}{16})^{\frac{1}{10}}\approx0.8298\). Then \(r = 1-0.8298 = 0.1702\).

Step3: Convert \(r\) to percentage

To convert \(r\) to a percentage, multiply by \(100\). So \(r = 0.1702\times100 = 17.02\%\).

Step4: Find the value in 2004

The time from 1991 to 2004 is \(t = 2004 - 1991=13\) years.
Using the formula \(y=a(1 - r)^t\) with \(a = 32000\), \(r=0.1702\), and \(t = 13\).
\(y=32000\times(1 - 0.1702)^{13}\).
\(y=32000\times(0.8298)^{13}\).
\((0.8298)^{13}\approx0.0997\).
\(y=32000\times0.0997 = 3190.4\approx3200\) (rounded to the nearest \(50\) dollars).

Answer:

A. \(r = 0.1702\)
B. \(r = 17.02\%\)
C. value \(=\$3200\)