Question 1153939

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

{{{P=4000}}}
{{{r=3}}}%={{{0.03}}}
{{{n=2}}}
{{{t=6}}} years

{{{A = 4000(1 + 0.03/2)^12}}}
{{{A = 4000(1 + 0.015)^12}}}
{{{A = 4000(1.015)^12}}}
{{{A = 4000(1.195618)}}}
{{{A = 4782.47}}}

Charles  will he have ${{{4782.47}}} after six years.