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A person invests 6000 dollars in a bank. The bank pays 6.5% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 8300 dollars? A=P\left(1+\frac{r}{n}\right)^{nt} A=P(1+ n r ​ ) nt

User Funktional
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2 Answers

9 votes

Final answer:

To find out how long the person must leave the money in the bank until it reaches $8300, we can use the formula for compound interest. By plugging in the given values and solving for t, we find that the person must leave the money in the bank for approximately 3.5 years.

Step-by-step explanation:

To find out how long the person must leave the money in the bank until it reaches $8300, we can use the formula for compound interest:



A = P(1 + r/n)^(nt)



Where:



A = the future value of the investment ($8300)



P = the principal amount ($6000)



r = the annual interest rate (6.5% or 0.065)



n = the number of times the interest is compounded per year (1, since it is compounded annually)



t = the number of years



Plugging in the given values, the equation becomes:



8300 = 6000(1 + 0.065/1)^(1 * t)



Now we can solve for t:



8300/6000 = (1.065)^t



1.38333 = 1.065^t



To solve for t, we take the logarithm of both sides:



log(1.38333) = log(1.065^t)



t = log(1.38333)/log(1.065)



Using a calculator, we find that t is approximately 3.5 years.

User Billeh
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5.2k points
7 votes

Answer: 5.2 years

Step-by-step explanation:

User AwfulHack
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5.1k points