Question

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.

Answer 1

To solve for the expected population in 2004:

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

Using compound interest formular to solve for the expected popupation:

Expected population = Amount

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

Hence the expected population in 2004 = 691,005