Answer:
A) $2,695.69
B) $2,713.02
C) See below.
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ I=P\left[\left(1+\frac{r}{n}\right)^{nt}-1\right]$\\\\where:\\\\ \phantom{ww}$\bullet$ $I =$ interest accrued \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $n =$ number of times interest is applied per year \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}[/tex]
Given:
Substitute the given values into the compound interest formula and solve for I:
[tex]\implies I=15000\left[\left(1+\dfrac{0.0475}{4}\right)^{4 \times 3.5}-1\right][/tex]
[tex]\implies I=15000\left[\left(1.011875\right)^{14}-1\right][/tex]
[tex]\implies I=15000(0.179712348...)[/tex]
[tex]\implies I=2695.68522...[/tex]
Therefore the interest earned is $2,695.69 to the nearest dollar.
[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Continuous Compounding Interest Formula}\\\\$ I=P\left(e^{rt}-1\right)$\\\\where:\\\\ \phantom{ww}$\bullet$ $I =$ interest accrued \\\phantom{ww}$\bullet$ $P =$ principal amount \\\phantom{ww}$\bullet$ $e =$ Euler's number (constant) \\\phantom{ww}$\bullet$ $r =$ annual interest rate (in decimal form) \\\phantom{ww}$\bullet$ $t =$ time (in years) \\\end{minipage}}[/tex]
Given:
Substitute the given values into the continuous interest formula and solve for I:
[tex]\implies I=15000\left(e^{0.0475 \times 3.5}-1\right)[/tex]
[tex]\implies I=15000\left(e^{0.16625}-1\right)[/tex]
[tex]\implies I=15000\left(0.180868281...\right)[/tex]
[tex]\implies I=2713.02422...[/tex]
Therefore the interest earned is $2,713.02 to the nearest dollar.
The difference between the interest earned for the two accounts is:
[tex]\implies 2713.02-2695.69=17.33[/tex]
Therefore, the principal will earn $17.33 more interest if the interest is compounded continuously rather than compounded quarterly.
