Question

Helpp pleaseeee

A car was valued at $26,000 in the year 1990. The value depreciated to 13,000 by the year 2003.

A) What was the annual rate of change between 1990 and 2003?
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 2006 ?

value = $ ------------- Round to the nearest 50 dollars.

Answer 1

A) To find the annual rate of change between 1990 and 2003, we can use the formula:

r = (V2/V1)^(1/n) - 1

where V1 is the initial value, V2 is the final value, and n is the number of years between the two values.

Plugging in the given values, we get:

r = (13000/26000)^(1/13) - 1

r = -0.0547

Therefore, the annual rate of change between 1990 and 2003 is -0.0547.

B) To express the rate of change as a percentage, we can multiply the result from part A by 100:

r = -0.0547 * 100

r = -5.47%

Therefore, the rate of change between 1990 and 2003 is a decrease of 5.47%.

C) To find the value of the car in the year 2006, we can use the formula:

V = V0 * (1 + r)^n

where V0 is the initial value, r is the annual rate of change, and n is the number of years between the initial value and the final value.

Since we want to find the value in 2006, which is 3 years after 2003, we have n = 3. Also, the initial value in 2003 was $13,000, so we have V0 = 13000. Finally, we can use the annual rate of change we found in part A:

V = 13000 * (1 - 0.0547)^3

V = $10,754.50

Therefore, the value of the car in the year 2006 would be approximately $10,755 (rounded to the nearest 50 dollars).