the table shows the account information of five investors. which of the following are true, assuming no…

Explanation:

Step1: Recall 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 $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (in decimal form), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years.

For Tara:

$P = 2100$, $r=0.045$, $n = 1$ (annually), $t = 6$. Using the compound - interest formula $A = P(1 + r)^t$, we have $A=2100(1 + 0.045)^6=2100\times1.045^6\approx2100\times1.30226\approx2734.75$.

For Ann:

$P = 4000$, $r = 0.05$, $n = 1$ (annually), $t = 12$. Using the compound - interest formula $A = P(1 + r)^t$, we have $A=4000(1 + 0.05)^{12}=4000\times1.05^{12}\approx4000\times1.79586\approx7183.44$.

For Nick:

$P = 2500$, $r = 0.03$, $n = 4$ (quarterly), $t = 8$. First, convert the formula to $A=P(1+\frac{r}{n})^{nt}$. Here, $\frac{r}{n}=\frac{0.03}{4}=0.0075$ and $nt=4\times8 = 32$. Then $A = 2500(1+0.0075)^{32}\approx2500\times1.27011\approx3175.28$.

For Steve:

$P = 3800$, $r = 0.035$, $n = 2$ (semi - annually), $t = 20$. First, $\frac{r}{n}=\frac{0.035}{2}=0.0175$ and $nt=2\times20 = 40$. Then $A = 3800(1 + 0.0175)^{40}\approx3800\times2.00159\approx7606.04$.

Answer:

A. After 6 years, Tara will have about $$2734.75$ (not $$2750.93$). B. After 15 years, there is no Lori in the table, so this option is incorrect. C. After 12 years, Ann will have about $$7183.44$ (not $$4788.33$). D. After 8 years, Nick will have about $$3175.28$ (not $$3177.17$). E. After 20 years, Steve will have about $$7606.04$ (not $$7629.00$). So, none of the options are correct.