Math Problem Statement

A car was valued at $39,000 in the year 1995. The value depreciated to $14,000 by the year 2003.

A) What was the annual rate of change between 1995 and 2003? r =  Round the rate of decrease to 4 decimal places.

Solution

To find the annual rate of change, we can assume that the value of the car depreciated exponentially. The formula for exponential decay is:

V(t)=V0×(1r)tV(t) = V_0 \times (1 - r)^t

Where:

  • V(t)V(t) is the value after tt years,
  • V0V_0 is the initial value,
  • rr is the annual rate of decrease,
  • tt is the number of years.

Given:

  • Initial value V0=39,000V_0 = 39,000 (in 1995),
  • Final value V(t)=14,000V(t) = 14,000 (in 2003),
  • t=20031995=8t = 2003 - 1995 = 8 years.

Now we can solve for rr by using the formula:

14,000=39,000×(1r)814,000 = 39,000 \times (1 - r)^8

First, isolate (1r)8(1 - r)^8 by dividing both sides by 39,000:

14,00039,000=(1r)8\frac{14,000}{39,000} = (1 - r)^8

Now, solve for (1r)(1 - r), and then for rr. I'll calculate this for you.The annual rate of decrease between 1995 and 2003 is approximately 0.1202, or 12.02%.

Would you like further details on this process, or do you have any other questions?

Here are five related questions to consider:

  1. How does the formula for exponential decay differ from linear depreciation?
  2. What would the car's value be in 2010 if the same rate of depreciation continued?
  3. How do you calculate depreciation using straight-line methods versus exponential?
  4. If the rate of decrease was 15%, what would the car's value have been in 2003?
  5. How can you calculate the rate of depreciation if given only two time points?

Tip: For exponential depreciation, the rate stays consistent over time, making it different from straight-line depreciation, which assumes a fixed amount lost each year.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Depreciation
Algebra

Formulas

V(t) = V_0 * (1 - r)^t
Annual rate of change formula: (V(t)/V_0)^(1/t) = 1 - r

Theorems

Exponential Decay Formula

Suitable Grade Level

Grades 9-12