Question

Compare the investment below to an investment of the same principal at the same rate compounded annually. ​principal: 4000​$​, annual​ interest: 3​%, interest​ periods: 6 , number of​ years: 14 Question content area bottom Part 1 After 14 ​years, the investment compounded periodically will be worth ​$ enter your response here more than the investment compounded annually. ​(Round to two decimal places as​ needed.

Answer 1

I think the answer is 6050.36? Don’t know 100% tho-

Answer 2

`~~~~~~ \stackrel{\textit{\LARGE Annually}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$4000\\ r=rate\to 3\%\to (3)/(100)\dotfill &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &14 \end{cases}`

`A = 4000\left(1+(0.03)/(1)\right)^(1\cdot 14) \implies A=4000(1.03)^(14)\implies \boxed{A \approx 6050.36} \\\\[-0.35em] ~\dotfill`

`~~~~~~ \stackrel{\textit{\LARGE BiMonthly}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$4000\\ r=rate\to 3\%\to (3)/(100)\dotfill &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{bimonthly, thus six} \end{array}\dotfill &6\\ t=years\dotfill &14 \end{cases}`

`A = 4000\left(1+(0.03)/(6)\right)^(6\cdot 14) \implies A=4000(1.005)^(84)\implies \boxed{A \approx 6081.48} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{periodic}{6081.48}~~ - ~~\stackrel{annual}{6050.36} ~~ \approx ~~ \text{\LARGE 31.12}`