pre calc question

Question

A car was valued at $28,000 in the year 1993. The value depreciated to $10,000 by the year 2003.

A) What was the annual rate of change between 1993 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 2007 ?

value = $   Round to the nearest 50 dollars.

Answer 1

10k/28k=5/14=e^10r

r= ln(5/14)/10= about -.102962

=about -.1030 rounded to 4 places

by 2007, another 4 years

FV=Poe^rt

FV = 28ke^14ln(5/14)= about 1.5 cents. that can't right

2003 $10,000

2004 $9000

2005 $8100

2006 $7290

2007 $6550

using 10% deoreciation

so it's a little less maybe about $6500 in 2007

Answer 2

This is a great question! First we will assume that the vehicle depreciates continuously, then we can use the equation A = P*e^(r*t).

Where A = final value

P = principal

r = rate

t = time elapsed

A = P*e^(rt)

Now we solve for rate.

A/P = e^(rt)

We take the natural logarithm of both sides.

ln(A/P) = r*t

r = ln(A/P)/t

For part B)

All you need to do is convert rate to percentage is multiply r*100 to get the percentage.

For part C) We use the rate we solved in part A and solve for the final value. We also change time to 15 years.

A = P*e^(r*t)

I hope that helps!