suppose you have $12,000 to invest. which of the two rates would yield the larger amount in 4 years: 11%…

Explanation:

Step1: Recall compound - interest formula for monthly compounding

The compound - interest formula for $n$ compounding periods per year is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times compounded per year, and $t$ is the number of years. Here, $P=$12000$, $r = 0.11$, $n = 12$ (monthly compounding), and $t = 4$. $A_1=12000(1 +\frac{0.11}{12})^{12\times4}$ $A_1=12000(1+\frac{0.11}{12})^{48}$ First, calculate $1+\frac{0.11}{12}=1+\ 0.009167=1.009167$. Then, $(1.009167)^{48}\approx1.5523$. So, $A_1 = 12000\times1.5523=$18627.6$.

Step2: Recall continuous - compounding formula

The continuous - compounding formula is $A=Pe^{rt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), and $t$ is the number of years. Here, $P = 12000$, $r=0.109$, and $t = 4$. $A_2=12000\times e^{0.109\times4}$ $A_2=12000\times e^{0.436}$ Since $e^{0.436}\approx1.5466$, then $A_2=12000\times1.5466=$18559.2$.

Answer:

11% compounded monthly