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A person invests 1000 dollars in a bank. The bank pays 6% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 2100 dollars?

A person invests 1000 dollars in a bank The bank pays 6 interest compounded annually To the nearest tenth of a year how long must the person leave the money in class=

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Solution

For this case we have the following information given:

A= 1000

r= 0.06

n = 1

A= 2100

So we can set up the following formula:

[tex]2100=1000(1+\frac{0.06}{1})^t[/tex]

We can solve for t on this way:

[tex]\ln (\frac{2100}{1000})=t\cdot\ln (1.06)[/tex]

Solving for t we got:

[tex]t=\frac{\ln (2.1)}{\ln (1.06)}=12.73[/tex]

Rounded to the nesrest tenth we got:

12.7 years

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