The population of a town in 2007 is 113, 505
and is increasing at a rate of 1.2% per year.
What will the population be in 2012?

What number will you fill in for a to solve the
equation?
(Hint: For this one, how many years
after 2007 is 2012?

Respuesta :

from 2007 to 2012 is only 5 years, so we can see this as a compound interest with a rate of 1.2% per annum for 5 years, so

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$113505\\ r=rate\to 1.2\%\to \frac{1.2}{100}\dotfill &0.012\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=113505\left(1+\frac{0.012}{1}\right)^{1\cdot 5}\implies A=113505(1.012)^5\implies A\approx 120481[/tex]

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