Question

You deposit $2000 into a savings account giving 7% interest compounded only at the end of the year. To nearest dollar, what is your end-of-year balance?

Answer 1

To calculate the end-of-year balance, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A = the end-of-year balance
P = the principal amount (initial deposit)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded in a year
t = the time (in years)

In this case, P = $2000, r = 7% = 0.07, n = 1 (compounded annually), and t = 1 (one year).

Plugging these values into the formula, we get:

A = 2000(1 + 0.07/1)^(1*1)
A ≈ $2,140.00

Therefore, to the nearest dollar, the end-of-year balance is $2,140.

Answer 2

well, you'd have the 2000 plus the 7% :|

`\begin{array}c \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{7\% of 2000}}{\left( \cfrac{7}{100} \right)2000}\implies 140~\hfill \underset{ new~balance }{\stackrel{ 2000~~ + ~~140 }{\text{\LARGE 2140}}}`