The equation for this case would be:
[tex]\begin{gathered} P=S\cdot(1+r)^n \\ P=\text{ final value, r= rate of increase, n=time in years, S= initial value} \end{gathered}[/tex]
Replacing the values, we have:
[tex]\begin{gathered} P=39000\cdot(1+0.055)^6 \\ P=39000\cdot(1.055)^6\text{ (Adding)} \\ P=39000\cdot(1.379)\text{ (Raising 1.055 to the power of 6)} \\ P=53774.9\text{ (Multiplying)} \\ \text{The answer is 53,775 rounded to the nearest dollar, which is the option }D \end{gathered}[/tex]
