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A person invests 7500 dollars in a bank. The bank pays 4% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 12800 dollars?

Respuesta :

nkirotedoreen46

Answer:

13.6 years

Step-by-step explanation:

From the question;

  • Money invested (Principal) is 7500 dollars
  • Rate of interest is 4% compounded annually
  • Amount, (money accrued ) is 12800 dollars

We are needed to determine the time it took for the money to reach the given amount;

  • To solve the question we need to know the compound interest formula, that is;

[tex]A=P(1-\frac{r}{100})^n[/tex]

Where n is the interest periods;

  • Therefore, substituting the variables with the corresponding values, we can determine n.

Therefore;

[tex]12800=7500(1-\frac{4}{100})^n[/tex]

Dividing both sides by 7500, we get;

[tex]1.7067=(1-\frac{4}{100})^n[/tex]

[tex]1.7067=(1.04)^n[/tex]

Introducing logarithms on both sides;

[tex]log1.7067=log(1.04)^n\\nlog1.04=log1.7067\\n=\frac{log1.7067}{log1.04} \\n=13.63[/tex]

Thus, n=13.6 years

Thus, it would take 13.6 years for the invested money to accumulate to 12800 dollars

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