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Joshua is going to invest $9,000 and leave it in an account for 5 years. Assuming theinterest is compounded continuously, what interest rate, to the nearest tenth of apercent, would be required in order for Joshua to end up with $12,500?

Sagot :

Let r be the percent annual interest rate of the account. Since $9000 are left for 5 years, for an outcome of $12,500, then:

[tex]9000\times(1+\frac{r}{100})^5=12,500[/tex]

Divide both sides by 9000:

[tex](1+\frac{r}{100})^5=\frac{12500}{9000}=\frac{25}{18}[/tex]

Take the 5th root to both sides:

[tex]\begin{gathered} 1+\frac{r}{100}=\sqrt[5]{\frac{25}{18}} \\ \Rightarrow\frac{r}{100}=\sqrt[5]{\frac{25}{18}}-1 \\ \Rightarrow r=100(\sqrt[5]{\frac{25}{18}}-1) \end{gathered}[/tex]

Use a calculator to find the decimal expression for r:

[tex]r=6.790716585\ldots[/tex]

Therefore, to the nearest tenth:

[tex]r=6.8[/tex]

This means that Joshua would need to invest his money on a 6.8% annual interest account.

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