We will use the equation [tex]A(t)=P(1+ \frac{r}{n})^{(n)(t)} [/tex], where P is the initial amount invested, r is the interest rate in decimal form, n is the number of times in a year the money is compounded, and t is the number of years the money will be invested. Our P = 300, r = .04, n = 12 (there are 12 months in a year), and t = 8. Filling in accordingly, [tex]A(t)=300(1+ \frac{.04}{12})^{(12)(8)} [/tex]. Simplifying what we can gives us [tex]A(t)=300(1+.00333333)^{96}[/tex]. Doing that addition inside the parenthesis and then raising that number to the 96th power gives us A(t) = 300(1.376395075) so A(t) = $412.92, choice B above.
