The rule of the interest compounded continuously is
[tex]A=Pe^{rt}[/tex]A is the new amount
P is the initial amount
r is the rate of interest in decimal
t is the time
The rule of the compounded monthly interest is
[tex]A=P(1+\frac{r}{12})^{12t}[/tex]A is the new amount
P is the initial amount
r is the interest rate in decimal
t is the time
Since the account pays 3.7% interest compounded continuously, then
[tex]r=\frac{3.7}{100}=0.037[/tex][tex]\begin{gathered} A=Pe^{0.037(1)} \\ A=P(1.037693021 \\ A=1.037693021P \end{gathered}[/tex]Since the other account pays 3.75% compounded monthly, then
[tex]r=\frac{3.75}{100}=0.0375[/tex][tex]\begin{gathered} A=P(1+\frac{0.0375}{12})^{12(1)} \\ A=P(1.038151293) \\ A=1.038151293P \end{gathered}[/tex]Since the value of A of the 2nd account is greater than the value of A in the 1st account, then
The better is the 2nd account
