Question

An investment of $7,000 is deposited into an account in which interest is compounded continuously. Complete the table by filling in the amounts to which the investment grows at the indicated times. (Round your answers to the nearest cent.)
r = 3%

Over a 6 year period

Answer 1

interest compounded continuously

\[Ae^{rt}\]

Answer 2

where A = your principle, starting value invested. and r is your interest rate, and t is the amount of time in years. usually t would be the number of times your interest is compounded but because it's compounded continuously t = the number of years.

Answer 3

\[7.0*10^{3}*(e^{0.03*6}) = ? \]

Answer 4

The way I understand this too is that the interest, compounded continuously can be derived from this formula for compound interest.

\[P*(1+\frac{ r }{ n })^{nt}\]

Answer 5

\[\large \color{black}{\begin{align}
\begin{array}{|c|c|c|} \hline
Year & Amount & Interest \ \\ \hline
1&7000&7000*0.03 \\ \hline
2&7210& 7210*0.03 \\ \hline
3&7426.3 & 7426.3*0.03 \\ \hline
4&7649.089 & 7649.089*0.03 \\ \hline
5&7878.56 & 7878.562*0.03 \\ \hline
6&8114.91 & 8114.91*0.03 \\ \hline
Final Amount & 8358.366 \\ \hline
\end{array}
\end{align}}\]
`Where Amount for year (N+1) = Amount of year N+Interest of year N`