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A car was valued at $27,000 in the year 1992. The value depreciated to $ the year 2006. A) What was the annual rate of change between 1992 and 2006? Round the rate of decrease to 4 decimal places. 7 = B) What is the correct answer to part A written in percentage form? T = %. C) Assume that the car value continues to drop by the same percentage. Y

A car was valued at $27,000 in the year 1992. The value depreciated to $ the year 2006. A) What was the annual rate of change between 1992 and 2006? Round the rate of decrease to 4 decimal places. 7 = B) What is the correct answer to part A written in percentage form? T = %. C) Assume that the car value continues to drop by the same percentage. Y

Elementary Algebra
Elementary Algebra
17th Edition
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:Lynn Marecek, MaryAnne Anthony-Smith
Chapter6: Polynomials
Section6.2: Use Multiplication Properties Of Exponents
Problem 168E: Depreciation Once a new car is driven away from the dealer, it begins to lose value. Each year, a...
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### Depreciation of a Car's Value: A Mathematical Approach #### Problem Statement A car was valued at $27,000 in the year 1992. The value depreciated to $15,000 by the year 2006. #### Questions **A) What was the annual rate of change between 1992 and 2006?** - \( r = \) ______________ *(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 2010?** - Value = $ ___________ *(Round to the nearest 50 dollars.)* ### Explanation and Solution 1. **Annual Rate of Change Calculation** To estimate the annual rate of depreciation, we use the formula for the exponential decay: \[ V(t) = V_0 \times (1 - r)^t \] Where: - \( V(t) \) is the value at time \( t \) - \( V_0 \) is the initial value - \( r \) is the annual rate of depreciation - \( t \) is the number of years Given: - \( V_0 = \$27,000 \) - \( V(14) = \$15,000 \) (from 1992 to 2006, \( t = 14 \) years) Plugging the given values into the formula: \[ 15,000 = 27,000 \times (1 - r)^{14} \] Dividing both sides by 27,000: \[ \left(\frac{15,000}{27,000}\right) = (1 - r)^{14} \] \[ \left(\frac{5}{9}\right) = (1 - r)^{14} \] Taking the 14th root of both sides: \[ \left(\frac{5}{9}\right)^{\frac{1}{14}} = 1 - r \] \[ 1 - r = 0.961271 \] \[ r = 1 - 0.961271 \] \[ r = 0.038729 \] So, the annual rate of change \( r \) is approximately \( 0.
Transcribed Image Text:### Depreciation of a Car's Value: A Mathematical Approach #### Problem Statement A car was valued at $27,000 in the year 1992. The value depreciated to $15,000 by the year 2006. #### Questions **A) What was the annual rate of change between 1992 and 2006?** - \( r = \) ______________ *(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 2010?** - Value = $ ___________ *(Round to the nearest 50 dollars.)* ### Explanation and Solution 1. **Annual Rate of Change Calculation** To estimate the annual rate of depreciation, we use the formula for the exponential decay: \[ V(t) = V_0 \times (1 - r)^t \] Where: - \( V(t) \) is the value at time \( t \) - \( V_0 \) is the initial value - \( r \) is the annual rate of depreciation - \( t \) is the number of years Given: - \( V_0 = \$27,000 \) - \( V(14) = \$15,000 \) (from 1992 to 2006, \( t = 14 \) years) Plugging the given values into the formula: \[ 15,000 = 27,000 \times (1 - r)^{14} \] Dividing both sides by 27,000: \[ \left(\frac{15,000}{27,000}\right) = (1 - r)^{14} \] \[ \left(\frac{5}{9}\right) = (1 - r)^{14} \] Taking the 14th root of both sides: \[ \left(\frac{5}{9}\right)^{\frac{1}{14}} = 1 - r \] \[ 1 - r = 0.961271 \] \[ r = 1 - 0.961271 \] \[ r = 0.038729 \] So, the annual rate of change \( r \) is approximately \( 0.
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