Each investment matures in 3 years. The interest compounds annually.
1 answer:
bearing in mind that 4¾ is simply 4.75.
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%5C%24600%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20n%3D%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%20%5Ctextit%7Bannually%2C%20thus%20once%7D%20%5Cend%7Barray%7D%5Cdotfill%20%261%5C%5C%20t%3Dyears%5Cdotfill%20%263%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D600%5Cleft%281%2B%5Cfrac%7B0.05%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%203%7D%5Cimplies%20A%3D600%281.05%29%5E3%5Cimplies%20A%3D694.575%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
well, the interest for each is simply A - P
695.575 - 600 = 95.575.
862.032 - 750 = 112.032.
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