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The yearly depreciation rate for a certain vehicle is modeled by \(r=1-\left(\frac{V}{C}\right)^{1 / n}\), where \(V\) is the value of the car after \(n\) years, and \(C\) is the original cost. a. Determine the depreciation rate for a car that originally cost \(\$ 18,000\) and is worth \(\$ 12,000\) after 3 yr. Round to the nearest tenth of a percent. b. Determine the original cost of a truck that has a yearly depreciation rate of \(15 \%\) and is worth \(\$ 11,000\) after 5 yr. Round to the nearest \(\$ 100\).

Short Answer

Expert verified
a. The depreciation rate is 12.6%. b. The original cost of the truck is approximately \$ 24,800.

Step by step solution

01

Understand the formula

The depreciation rate formula given is \[ r = 1 - \left( \frac{V}{C} \right)^{1/n} \]Where: - \(r\) is the yearly depreciation rate, - \(V\) is the value of the vehicle after \(n\) years, - \(C\) is the original cost, - \(n\) is the number of years.
02

Substitute values for part a

Given \(C = 18000\), \(V = 12000\), and \(n = 3\). Substitute these values into the formula:\[ r = 1 - \left( \frac{12000}{18000} \right)^{1/3} \]
03

Simplify the fraction

Simplify \( \frac{12000}{18000} \) to get \( \frac{2}{3} \):\[ r = 1 - \left( \frac{2}{3} \right)^{1/3} \]
04

Calculate the cube root

Determine the cube root of \( \frac{2}{3} \):\[ \left( \frac{2}{3} \right)^{1/3} \approx 0.8736 \]
05

Compute the depreciation rate

Subtract the cube root from 1:\[ r = 1 - 0.8736 \approx 0.1264 \]Convert to a percentage and round to the nearest tenth of a percent:\[ r \approx 12.6\text{%} \]
06

Set up equation for part b

Given \(r = 0.15\), \(V = 11000\), and \(n = 5\), rearrange the formula to solve for \(C\):\[ 0.15 = 1 - \left( \frac{11000}{C} \right)^{1/5} \]
07

Isolate the fractional term

Isolate the fractional term on one side of the equation:\[ 0.85 = \left( \frac{11000}{C} \right)^{1/5} \]
08

Raise both sides to the power of 5

Remove the exponent by raising both sides to the power of 5:\[ 0.85^5 = \frac{11000}{C} \]Calculate \(0.85^5\):\[ 0.85^5 \approx 0.4437 \]
09

Solve for C

Solve for \(C\) by rearranging the equation:\[ C = \frac{11000}{0.4437} \approx 24800 \]Round to the nearest \$ 100:\[ C \approx 24800 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vehicle Depreciation
Depreciation is the reduction in the value of an asset over time, often due to wear and tear. For vehicles, depreciation is a significant factor to consider when buying, selling, or valuing cars and trucks. The depreciation rate indicates how much the value of the vehicle decreases each year.
The given formula for this problem is \[ r = 1 - \bigg( \frac{V}{C} \bigg)^{1/n} \ \text{where:} \ r \text{ is the yearly depreciation rate} \ V \text{ is the value of the vehicle after } n \text{ years} \ C \text{ is the original cost} \ n \text{ is the number of years} \]
Using this formula, you can find the depreciation rate if you know the original cost and the value after a certain number of years. This helps you understand how quickly a vehicle loses its value over time, which is crucial for both buyers and sellers.
Exponential Decay
Exponential decay is a process by which a quantity decreases at a rate proportional to its current value. In the context of vehicle depreciation, it means that the value of the car decreases by a certain percentage each year.
This concept is mathematically represented in the given formula. The term \( \big( \frac{V}{C} \big)^{1/n} \) represents the fraction of the original cost that the vehicle maintains each year. By subtracting this from 1, we get the annual depreciation rate.
Think of it this way: if a car loses 20% of its value each year, it follows an exponential decay pattern. Each year, the car's value is 80% of the value from the previous year. The formula helps quantify this decay, making it easier to predict future values. Exponential decay is a common concept in finance and helps in understanding the diminishing returns of valuable assets over time.
Financial Mathematics
Financial mathematics involves the application of mathematical methods to solve problems in finance, such as calculating depreciation rates. Using mathematical formulas like the one given in the problem helps in making informed decisions about investments, purchases, and valuations.
In this exercise, we applied financial mathematics to determine how much a vehicle's value will decrease over a specified period. This involves substituting known values into the formula and performing algebraic manipulations to solve for the desired variable. For instance, we had to isolate the fractional term and take the cube root to find the depreciation rate in part a.
Financial mathematics is crucial for areas like budgeting, forecasting, and financial planning. It provides the tools needed to analyze and interpret financial data effectively. Understanding these concepts can help you navigate various financial scenarios, from buying a car to managing long-term investments.

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