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Travka [436]
3 years ago
12

A car was valued at $39,000 in the year 1995. The value depreciated to $11,000 by the year 2003.

Mathematics
1 answer:
denis23 [38]3 years ago
5 0

Answer:

14.6328% , $5836.03

Step-by-step explanation:

Here we are going to use the formula

A_{0}(1-r)^n = A_{n}

A_{0} = 39000

r=?

A_{8} = 11000

n=8

Hence

39000(1-r)^8 = 11000

(1-r)^8 = \frac{11000}{39000}

(1-r)^8 = 0.2820

(1-r) = 0.2820^{\frac{1}{8}

(1-r) = 0.2820^{0.125}

(1-r) = 0.8536

(1-0.8536=r

r = 0.1463

Hence r= 0.1463

In percentage form r = 14.63%

Now let us see calculate the value of car in 2003 that is after 12 years

we use the main formula again

A_{0}(1-r)^n = A_{n}

A_{0} = 39000

r=0.1463

A_{12} = ?

n=12

39000(1-0.14634)^{(12} = A_{12}

39000(0.8536)^{12} = A_{12}

39000*0.1497 = A_{12}

A_{12}=5840.34

Hence the car's value will be depreciated to $5840.34 (approx) by 2003.

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djverab [1.8K]
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Step-by-step explanation:

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