hello
this question involves compound interest and we can easily solve this using the formula for it
first of all, let's get our data out
principal (p) = 4000
rate = 3% = 0.03
time (t) = 6 years
number of times compounded in a year (n) = 1
[tex]\begin{gathered} c\mathrm{}p=p(1+\frac{r}{n})^{nt} \\ c\mathrm{}p=4000(1+\frac{0.03}{1})^{(1\times6)} \\ c\mathrm{}p=4000(1+0.03)^6 \\ c\mathrm{}p=4000\times1.03^6 \\ c\mathrm{}p=4776.209 \end{gathered}[/tex]
the interest compounded in 6 years is equal to $4776.209
