Question

Help please i need the answers in order please

A car was valued at $42,000 in the year 1994. The value depreciated to $11,000 by the year 2006.

A) What was the annual rate of change between 1994 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.

Answer 1

A) To find the annual rate of change, we can use the formula:

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

where:

V1 = initial value ($42,000)

V2 = final value ($11,000)

n = number of years (2006 - 1994 = 12)

Plugging in the values, we get:

r = (11000/42000)^(1/12) - 1 ≈ -0.1135

Therefore, the annual rate of change between 1994 and 2006 is approximately -0.1135.

B) To convert the rate of change to a percentage, we can multiply by 100 and add a percent sign:

r = -0.1135 × 100% ≈ -11.35%

Therefore, the correct answer to part A written in percentage form is approximately -11.35%.

C) Assuming the car value continues to drop by the same percentage, we can use the formula for exponential decay:

V = V0 * (1 - r)^t

where:

V0 = initial value ($11,000 in 2006)

r = annual rate of change (-0.1135)

t = number of years (2010 - 2006 = 4)

Plugging in the values, we get:

V = 11000 * (1 - (-0.1135))^4 ≈ $6,250

Therefore, the value of the car in the year 2010 would be approximately $6,250, rounded to the nearest 50 dollars.