Answer:
The $3485.52 money did Meg have at the end of the account term.
Step-by-step explanation:
Formula for compounded monthly
[tex]Amount = P(1+\frac{r}{365})^{365n}[/tex]
Where P is the principle , r is the rate of interest in the decimal form and n is the number of years.
As given
Meg deposited a $3,000 bonus check in a new savings account. The account has an interest rate of 3% for 5 years.
Principle = $3000
3% is written in the decimal form
[tex]= \frac{3}{100}[/tex]
= 0.03
Time = 5 years
Put in the formula
[tex]Amount = 3000(1+\frac{0.03}{365})^{365\times 5}[/tex]
[tex]Amount = 3000(1+\frac{0.03}{365})^{1825}[/tex]
[tex]Amount = 3000(1+\frac{0.03}{365})^{1825}[/tex]
[tex]Amount = 3000(1+ 0.0000822)^{1825}[/tex]
[tex]Amount = 3000(1.0000822)^{1825}[/tex]
[tex]Amount = 3000\times 1.16184[/tex]
Amount = $ 3485.52
Therefore the $ 3485.52 money did Meg have at the end of the account term.
