suppose you have $13,000 to invest. which of the two rates would yield the larger amount in 3 years: 7%…

Explanation:

Step1: Calculate amount for 7% compounded monthly

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$. Here, $P=$13000$, $r = 0.07$, $n = 12$ (monthly compounding), and $t = 3$. [ \begin{align*} A&=13000\left(1+\frac{0.07}{12}\right)^{12\times3}\ &=13000\left(1+\frac{0.07}{12}\right)^{36}\ \end{align*} ] First, calculate $\frac{0.07}{12}\approx0.005833$. Then $1+\frac{0.07}{12}\approx1.005833$. $(1.005833)^{36}\approx1.2329$. $A = 13000\times1.2329=$16027.7$

Step2: Calculate amount for 6.91% compounded continuously

The continuous - compounding formula is $A = Pe^{rt}$. Here, $P = 13000$, $r=0.0691$, and $t = 3$. [ \begin{align*} A&=13000e^{0.0691\times3}\ &=13000e^{0.2073}\ \end{align*} ] Since $e^{0.2073}\approx1.2309$. $A=13000\times1.2309=$15991.7$

Answer:

7% compounded monthly