Question

At the beginning of year 1, paolo invests $500

Answer 1

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 500\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annuall, thus once} \end{array}\to &1\\ t=years\to &t \end{cases} \\\\\\ A=500\left(1+\frac{0.04}{1}\right)^{1\cdot t}\implies A=500(1+0.04)^t \\\\\\ \textit{after 5 years }t=5\qquad A=500(1.04)^5$

the example on your picture uses A(n) and n = years, but is pretty much the same, in this case is t = years.

Answer 2

Answer:

The correct option is C.

Step-by-step explanation:

Principal Amount = $500

Rate of Interest = 4%

                          = 0.04

Time = 4 years

n = 1

$Amount = 500\times(1+0.04)^{4}\\\\\implies Amount = 500\times 1.1699\\\\\implies Amount = \ 584.93$

Hence, The explicit formula which can be used to find the amount of the money in the account :

$A(n)=500\cdot (1+0.04)^{(n-1)}$

Therefore, The correct option is C.