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fiasKO [112]
4 years ago
14

A new car is purchased for 23900 dollars. The value of the car depreciates at 14.25% per year. To the nearest tenth of a year, h

ow long will it be until the value of the car is 6100 dollars?
Mathematics
2 answers:
svp [43]4 years ago
7 0

Answer:

The value of the car will be 6100 dollars in 8.8 years

Step-by-step explanation:

Present value of car = $23900

The value of the car depreciates at 14.25% per year

Let x be no. of years in which  the value of the car becomes 6100 dollars

Formula: N(t)=N_0(1-r)^t

Substitute the values :

6100=23900(1-\frac{14.25}{100})^x\\\frac{6100}{23900}=(1-\frac{14.25}{100})^x\\x=8.8

Hence the value of the car will be 6100 dollars in 8.8 years

sleet_krkn [62]4 years ago
5 0

Answer:

The correct answer is 8.9

Step-by-step explanation:

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(1/6)2 has a value of 1/36

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Using arithmetic sequence concepts, it is found that the common difference is of 0.25.

<h3>What is an arithmetic sequence?</h3>

In an arithmetic sequence, the difference between consecutive terms is always the same, called common difference d.

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