A new car is purchased for 16000 dollars. The value of the car depreciates at 6.25% per year. To the nearest tenth of a year, how long will it be until the value of the car is 11500 dollars?

5.1 years

Step-by-step explanation:

the formula for fixed rate depreciation is

Value after t years = initial value ( 1 - r )^t

where r = depreciation rate = 6.25% = 0.0625

t = length of time in years (we are asked to find this

Value after t years = given as $11,500

initial Value = given as $16,000

Substituting this into equation

11,500 = 16,000 (1 - 0.0625) ^t

11,500 = 16,000 (0.9375) ^t  (divide both sides by 16,000 and rearrange)

0.9375^t = 11,500 / 16000

0.9375^t = 0.71875  (taking log of both sides)

t log 0.9375 = log 0.71875

t =   log 0.71875 / log 0.9375  (use calculator)

t = 5.12

t = 5.1 years (nearest tenth)

it will take 5.1 years

Step-by-step explanation:

We would apply the formula for exponential decay which is expressed as

A = P(1 - r)^ t

Where

A represents the value of the car after t years.

t represents the number of years.

P represents the value of the car.

r represents rate of decay.

From the information given,

A = $11500

P = $16000

r = 6.25% = 6.25/100 = 0.0625

Therefore

11500 = 16000(1 - 0.0625)^t

11500/16000( = 0.9375)^t

0.71875 = (0.9375)^t

Taking log of both sides to base 10

Log 0.71875 = tlog 0.9375

- 0.143 = - 0.028t

t = - 0.143/- 0.028

t = 5.1 years

answer: 1) twenty-seven times y

              2) twenty-seven multiplied by y