Complete the table by finding the balance $A$ when $P$ dollars is invested at rate $r$ for $t$ years and compounded $n$ ti year. (Round your answers to the nearest cent.) \[ P=\$ 2100, r=5.5 \%, t=12 \text { years } \] \begin{tabular}{|c|c|} \hline$n$ & $A$ \\ \hline 1 & $\$$ \\ \hline 2 & $\$$ \\ \hline 4 & $\$$ \\ \hline 12 & $\$$ \\ \hline 365 & $\$$ \\ \hline Continuous & $\$$ \\ \hline \end{tabular} Need Help? Read It Watch It

Step 1 ：The problem is asking to find the balance $A$ when $P$ dollars is invested at rate $r$ for $t$ years and compounded $n$ times a year. The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, 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 ：In this case, $P = 2100$, $r = 5.5\% = 0.055$, and $t = 12$ years. We need to calculate $A$ for different values of $n$ (1, 2, 4, 12, 365, and continuous).

Step 3 ：For the continuous case, the formula for continuous compounding is $A = Pe^{rt}$, where $e$ is the base of the natural logarithm (approximately equal to 2.71828).

Step 4 ：Let's calculate $A$ for each value of $n$.

Step 5 ：The balances are approximately \$3992.54, \$4027.01, \$4044.84, \$4056.94, \$4062.86, and \$4063.06 respectively.

Step 6 ：Final Answer: The balances $A$ when $P = \$2100$ is invested at rate $r = 5.5\%$ for $t = 12$ years and compounded $n$ times a year are approximately \$3992.54, \$4027.01, \$4044.84, \$4056.94, \$4062.86, and \$4063.06 respectively. For continuous compounding, the balance is approximately \$4063.06.