Step 1
State the compound interest formula
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]where;
[tex]\begin{gathered} P=\text{ \$}9500 \\ r=\frac{4.5}{100}=0.045 \\ n=4 \\ A=\text{ \$22100} \end{gathered}[/tex]Step 2
Find how long must the person leave the money in the bank until it reaches 22100 dollars
[tex]\begin{gathered} 22100=9500(1+\frac{0.045}{4})^{4\times t} \\ 22100=9500(\frac{809}{800})^{4t} \\ \frac{9500\left(\frac{809}{800}\right)^{4t}}{9500}=\frac{22100}{9500} \\ \left(\frac{809}{800}\right)^{4t}=\frac{221}{95} \\ 4t\ln \left(\frac{809}{800}\right)=\ln \left(\frac{221}{95}\right) \\ t=\frac{\ln \left(\frac{221}{95}\right)}{4\ln \left(\frac{809}{800}\right)} \\ t=18.86724 \\ t\approx18.9\text{ years to the nearest tenth of a year} \end{gathered}[/tex]Answer;
[tex]t=18.9\text{ years to the nearest tenth of a year}[/tex]