Step 1 :The formula for compound interest is given by: \(A = P(1 + r/n)^{nt}\) where: \(A\) is the amount of money accumulated after n years, including interest, \(P\) is the principal amount (the initial amount of money), \(r\) is the annual interest rate (in decimal), \(n\) is the number of times that interest is compounded per year, and \(t\) is the time the money is invested for in years.
Step 2 :For annually compounding, we have \(n = 1\). So, \(A = 13000(1 + 0.039/1)^{1*16} = 13000(1.039)^{16} = \$23,759\) (rounded to the nearest whole number).
Step 3 :For quarterly compounding, we have \(n = 4\). So, \(A = 13000(1 + 0.039/4)^{4*16} = 13000(1.00975)^{64} = \$24,051\) (rounded to the nearest whole number).
Step 4 :For monthly compounding, we have \(n = 12\). So, \(A = 13000(1 + 0.039/12)^{12*16} = 13000(1.00325)^{192} = \$24,119\) (rounded to the nearest whole number).
Step 5 :For daily compounding, we have \(n = 365\). So, \(A = 13000(1 + 0.039/365)^{365*16} = 13000(1.00010684931507)^{5840} = \$24,152\) (rounded to the nearest whole number).
Step 6 :For continuously compounding, the formula is \(A = Pe^{rt}\). So, \(A = 13000e^{0.039*16} = \$24,158.54\) (rounded to the nearest cent).
Step 7 :So, the completed table is: \begin{tabular}{|l|l|c|c|} \hline & Compounding Option & $n$ Value & Result \\ \hline (a) & Annually & $n=1$ & $\boxed{23,759}$ \\ \hline (b) Quarterly & $n=4$ & $\boxed{24,051}$ \\ \hline (c) Monthly & $n=12$ & $\boxed{24,119}$ \\ \hline (d) Daily & $n=365$ & $\boxed{24,152}$ \\ \hline (e) Continuously & Not Applicable & $\boxed{24,158.54}$ \\ \hline \end{tabular}