Respuesta :
The population at the end of the year is
(year-end population) = (beginning population) + 6.5%×(beginning population)
= (beginning population)×(1 + 0.065)
That same multiplier applies each year for 9 years. We know that we can use an exponent to describe repeated multiplication, so the population after 9 years will be ...
(ending population) = (beginning population)×(1.065)⁹
After 9 years, the population will be 350,000×1.065⁹ ≈ 616,900.
(year-end population) = (beginning population) + 6.5%×(beginning population)
= (beginning population)×(1 + 0.065)
That same multiplier applies each year for 9 years. We know that we can use an exponent to describe repeated multiplication, so the population after 9 years will be ...
(ending population) = (beginning population)×(1.065)⁹
After 9 years, the population will be 350,000×1.065⁹ ≈ 616,900.
to the risk of sounding redundant.
[tex]\bf \qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\to &350000\\ r=rate\to 6.5\%\to \frac{6.5}{100}\to &0.065\\ t=\textit{elapsed time}\to &9\\ \end{cases} \\\\\\ A=350000(1+0.065)^9\implies A\approx 616899.63647 \\\\\\ \textit{which rounds up to }616900~folks[/tex]
[tex]\bf \qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\to &350000\\ r=rate\to 6.5\%\to \frac{6.5}{100}\to &0.065\\ t=\textit{elapsed time}\to &9\\ \end{cases} \\\\\\ A=350000(1+0.065)^9\implies A\approx 616899.63647 \\\\\\ \textit{which rounds up to }616900~folks[/tex]