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The population of Somewhere, USA was estimated to be 658,100 in 2003, with an expected increase of 5% per year. At the percent ofincrease given, what was the expected population in 2004? Round your answer to the nearest whole number.

Respuesta :

To solve for the expected population in 2004:

[tex]\begin{gathered} \text{Estimated population for 2003=658100} \\ \text{rate = 5 \%} \\ nu\text{mber of year = 1} \end{gathered}[/tex]

Using compound interest formular to solve for the expected popupation:

Expected population = Amount

[tex]\begin{gathered} A=p(1+\frac{r}{100})^n \\ A\text{ = 658100 (1+}\frac{5}{100})^1 \\ A=658100\text{ (1+0.05)} \\ A=658100(1.05) \\ A=691005 \end{gathered}[/tex]

Hence the expected population in 2004 = 691,005

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