if 44300 dollars is invested at an interest rate of 5 percent per year, find the value of the investment at…

Explanation:

Step1: Identify the compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$ for non - continuous compounding and $A = Pe^{rt}$ for continuous compounding, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $n$ is the number of times compounded per year, $t$ is the number of years, and $A$ is the amount of money accumulated after $n$ years, including interest. Given $P=$44300$, $r = 0.05$, and $t = 5$.

Step2: Calculate for annual compounding ($n = 1$)

Substitute $P = 44300$, $r=0.05$, $n = 1$, and $t = 5$ into $A = P(1+\frac{r}{n})^{nt}$. $A=44300(1 +\frac{0.05}{1})^{1\times5}=44300(1.05)^{5}\approx44300\times1.27628\approx56539.1$.

Step3: Calculate for semiannual compounding ($n = 2$)

Substitute $P = 44300$, $r = 0.05$, $n = 2$, and $t = 5$ into $A = P(1+\frac{r}{n})^{nt}$. $A=44300(1+\frac{0.05}{2})^{2\times5}=44300(1 + 0.025)^{10}\approx44300\times1.2800845\approx56607.74$.

Step4: Calculate for monthly compounding ($n = 12$)

Substitute $P = 44300$, $r = 0.05$, $n = 12$, and $t = 5$ into $A = P(1+\frac{r}{n})^{nt}$. $A=44300(1+\frac{0.05}{12})^{12\times5}=44300(1+\frac{0.05}{12})^{60}\approx44300\times1.2833587\approx56852.79$.

Step5: Calculate for daily compounding ($n = 365$)

Substitute $P = 44300$, $r = 0.05$, $n = 365$, and $t = 5$ into $A = P(1+\frac{r}{n})^{nt}$. $A=44300(1+\frac{0.05}{365})^{365\times5}=44300(1+\frac{0.05}{365})^{1825}\approx44300\times1.2840254\approx56882.32$.

Step6: Calculate for continuous compounding

Substitute $P = 44300$, $r = 0.05$, and $t = 5$ into $A = Pe^{rt}$. $A=44300e^{0.05\times5}=44300e^{0.25}\approx44300\times1.2840254\approx56882.32$.

Answer:

(a) $$56539.1$ (b) $$56607.74$ (c) $$56852.79$ (d) $$56882.32$ (e) $$56882.32$