Question

An investment of $450 increases at a rate of 6.5% per year. what is the growth factor, b?

Answer 1

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\ 450\\ r=rate\to 6.5\%\to \frac{6.5}{100}\dotfill &0.065\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years \end{cases}$

$\bf A=450\left(1+\frac{0.065}{1}\right)^{1\cdot t}\implies \underset{\textit{growth factor}}{A=450(\underset{\uparrow }{1.065})^t}$