Respuesta :
The Solution:
To find the annual interest rate that will help Jaquin to meet his goal, we shall use the formula below:
[tex]A=P+I=P(1+\frac{r}{100\alpha})^{n\alpha}[/tex]In this case,
[tex]\begin{gathered} A=\text{amount}=90000+900=\text{ \$90900} \\ P=\text{ amount to be invested= \$90000} \\ I=\text{interest}=\text{ \$900} \\ r=\text{ interst rate per year=?} \\ n=\text{ number of years = 3.6 years} \\ \alpha=\text{ number of periods per year=365 (since it is compounded daily)} \end{gathered}[/tex]Substituting these values in the above formula, we get
[tex]90900=90000(1+\frac{r}{(100\times365)})^{(3.6\times365)}[/tex][tex]90900=90000(1+\frac{r}{(36500)})^{1314}[/tex]Dividing both sides by 90000, we get
[tex]\begin{gathered} \frac{90900}{90000}=(1+\frac{r}{(36500)})^{1314} \\ \\ 1.01=(1+\frac{r}{36500})^{1314} \end{gathered}[/tex]Multiplying the power of both sides by 1/1314, we get
[tex]\begin{gathered} 1.01^{\frac{1}{1314}}=(1+\frac{r}{36500})^{1314\times\frac{1}{1314}} \\ \\ 1.000007573=1+\frac{r}{36500} \end{gathered}[/tex]Collecting the like terms, we get
[tex]\begin{gathered} 1.000007573-1=\frac{r}{36500} \\ \\ 0.000007573=\frac{r}{36500} \end{gathered}[/tex]Cross multiplying, we get
[tex]r=0.000007573\times36500=0.2764145\approx0.28\text{ \%}[/tex]Therefore, the correct answer is 0.28%
