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N76 [4]
3 years ago
15

in the year 2001, a person bought a new car for $22000. for each consecutive year after that, the value of the car depreciated b

y 5%. how much would the car be worth in the year 2005, to the nearest hundred dollars
Mathematics
1 answer:
igomit [66]3 years ago
7 0

Answer:

To the nearest hundred dollars, the car will be worth $17,900 by 2005

Step-by-step explanation:

Firstly, we need to write the depreciation equation

We have this as:

V = I(1 - r)^t

V is the present value which is what we want to calculate

I is the initial value, the amount the cad was bought which is $22,000

r is the rate of change which is 5% = 5/100 = 0.05

t is the time difference which is 2005-2001 = 4

Substituting all these into the depreciation equation, we have it that

V = 22,000(1 -0.05)^4

V = $17,919.1375

To the nearest hundred dollars, that would be;

$17,900

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kondor19780726 [428]

The tenth position is the one that goes after the decimal point.

To round a number, you have to take into account the following:

1. If the number that goes after the position we are going to round to is greater than 5, we round to the next number in that position.

2. If the number that goes after the position we are going to round to is less than 5, we round to the same number in that position.

In this case, the number that is on the tenth's position is 4. The number that is after this position is 1, which is less than 5, then we round the number in this positon to 4.

The rounded number would be:

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