question 3 $5000 are invested in a bank account at an interest rate of 9 percent per year. find the amount…

Explanation:

Step1: Recall compound - interest formulas

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$ for compounded $n$ times a year and $A = Pe^{rt}$ for continuous compounding, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $t$ is the number of years, and $n$ is the number of times compounded per year. Here, $P=$5000$, $r = 0.09$, and $t = 14$.

Step2: Calculate amount when compounded annually ($n = 1$)

Substitute $P = 5000$, $r=0.09$, $n = 1$, and $t = 14$ into $A = P(1+\frac{r}{n})^{nt}$. $A=5000(1 +\frac{0.09}{1})^{1\times14}=5000(1.09)^{14}\approx5000\times3.341208\approx16706.04$

Step3: Calculate amount when compounded quarterly ($n = 4$)

Substitute $P = 5000$, $r = 0.09$, $n = 4$, and $t = 14$ into $A = P(1+\frac{r}{n})^{nt}$. $A=5000(1+\frac{0.09}{4})^{4\times14}=5000(1 + 0.0225)^{56}\approx5000\times3.467894\approx17339.47$

Step4: Calculate amount when compounded monthly ($n = 12$)

Substitute $P = 5000$, $r = 0.09$, $n = 12$, and $t = 14$ into $A = P(1+\frac{r}{n})^{nt}$. $A=5000(1+\frac{0.09}{12})^{12\times14}=5000(1+0.0075)^{168}\approx5000\times3.512917\approx17564.59$

Step5: Calculate amount when compounded continuously

Substitute $P = 5000$, $r = 0.09$, and $t = 14$ into $A = Pe^{rt}$. $A=5000e^{0.09\times14}=5000e^{1.26}\approx5000\times3.526361\approx17631.81$

Answer:

$16706.04$ $17339.47$ $17564.59$ $17631.81$