Answered

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A person invests 8000 dollars in a bank. The bank pays 5.25% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 12700 dollars?

Sagot :

fieryanswererft

Answer:

she should save it for minimum 9.1 years.

Explanation:

  • compound interest formula: [tex]\sf A = P(1+\dfrac{r}{100} )^n[/tex]
  • where A is money earned, P is investing money, r is rate of interest, n is time.

solve:

[tex]\hookrightarrow \sf 12700 =8000 (1+\dfrac{5.25}{100} )^n[/tex]

change sides

[tex]\hookrightarrow \sf \dfrac{12700}{8000} = (1+\dfrac{5.25}{100} )^n[/tex]

take ln on both sides

[tex]\hookrightarrow \sf n\ln \left(1+\dfrac{5.25}{100}\right)=\ln \left(\dfrac{127}{80}\right)[/tex]

simplify

[tex]\hookrightarrow \sf n=\frac{\ln \left(\dfrac{127}{80}\right)}{\ln \left(\dfrac{105.25}{100}\right)}[/tex]

final answer

[tex]\hookrightarrow \sf n=9.03216[/tex]

rounding to nearest tenth

[tex]\hookrightarrow \sf n=9.1[/tex]

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