We have a compounded monthly interest, the formula for compounded interest is
[tex]A=P\mleft(1+\frac{i}{n}\mright)^{nt}[/tex]
The tricky part is the "n", when it's compounded monthly we must do n = 12.
The other variables are
t - time
i - rate per year
P - principal
A - accrued amount
Looking at the problem we can identify that
t = 3 years
i = 0.0525 per year
P = ?
A = $3350
Then, if we put it at the formula we get
[tex]\begin{gathered} A=P\mleft(1+\frac{i}{n}\mright)^{nt} \\ \\ 3350=P\mleft(1+\frac{0.0525}{12}\mright)^{12\cdot3} \\ \\ 3350=P(1.004375)^{36} \\ \\ 3350=P\cdot1.17017 \\ \\ P=\frac{3350}{1.17017} \\ \\ P=\$2862.81 \end{gathered}[/tex]
Jason borrowed $2862.81
