pre calc question 10

Question

A car was valued at $25,000 in the year 1992. The value depreciated to $10,000 by the year 2001.

A) What was the annual rate of change between 1992 and 2001? 

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 2004 ?

value = $   Round to the nearest 50 dollars.

Answer 1

A) C(t) = 25000(1 - r)^t. In 2001 t = 2001 - 1992 = 9, P(8) = 10000. So, 10000 = 25000(1 - r)⁹.

r = 1 - (10000/25000)^1/9 = 0.0968.

B) r = %9.68;

C) In 2004 t = 2004 - 1992 = 12; C(12) = 25000(1 - 0.0968)¹² = $7367.96 = $7350

Answer 2

Start with two-point form from Algebra 1.

(y - 1992) / (x - 2500) = (2001 - 1995) /. (10000 - 25000)

Convert to slope-interept form and answer.