Question

Maria deposited $1,000 into an account at this rate. How much money was in the account after 3 years , interest of 6.5%? (Assume that the account earns compound interest). Give your answer in fraction.

Answer 1

Step 1:

`\begin{gathered} \text{A = P(1 + r)}^(nt) \\ \\ \text{For annually, n = 1} \end{gathered}`

Step 2:

P = $1000

t = 3 years

r = 6.5% = 0.065

Step 3:

`\begin{gathered} A=P(1+r)^(nt) \\ A\text{ = 1000 }*(1+0.065)^(1*3) \\ A\text{ = 1000 }*(1.065)^3 \\ A\text{ = }1000\text{ }*\text{ (}(1065)/(1000))^3 \\ A\text{ = 1000 }*\text{ }(1207949625)/(1000000000) \\ A\text{ = }(1207949625)/(1000000) \\ A\text{ = \$}(9663597)/(8000) \end{gathered}`

Final answer

`A\text{ = \$}(9663597)/(8000)`

1207.950