A new car sells for $27,300. It exponentially depreciates at a rate of 6.1% to $22,100. How long did it take for the car to depreciate to this amount? Round your answer to the nearest tenth of a year. fill in the blank 1

Answers

Answer 1

Answer:

3

Step-by-step explanation:

27300 * (1-6.1%)^n = 22100,

n∈(3,4)

when n = 3,≈22603;

when n = 4,≈21224;

so choose n = 3.

Answer 2

Therefore, it takes  car to depreciate to this amount is [tex]3[/tex].

What is the simplification?

Simplification is reducing the expression/fraction/problem in a simpler form. It makes the problem easy with calculations and solving.

Here given that,

A new car sells for $[tex]27,300[/tex]. It exponentially depreciates at a rate of [tex]6.1[/tex]% to $[tex]22,100[/tex].

So, the equation is

[tex]27300(1-6.1)^n=22100\\\\[/tex]

when [tex]n[/tex] belongs to [tex](3,4)[/tex]

when [tex]n=3\\[/tex]

The equation would be

[tex]27300(1-6.1)^3=22100\\\\=22603[/tex]

When

[tex]n=4\\27300(1-6.1)^4=22100\\=21224[/tex]

So, the time it takes to depricate to the given amount is [tex]22603[/tex] when [tex]n=3[/tex].

Hence, it takes  car to depreciate to this amount is [tex]3[/tex].

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Answer:

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Answer:

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Answers

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1.

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Step-by-step explanation:

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