Solution:
The continuous compound interest is expressed as
[tex]\begin{gathered} P(t)=P_0\times e^{rt} \\ where \\ P(t)=value\text{ at time t} \\ P_0=principal\text{ amount} \\ r=annual\text{ interest rate} \\ t=length\text{ of time the interest is applied} \end{gathered}[/tex]
Given that
[tex]\begin{gathered} t=6 \\ r=3.5\%=0.035 \\ P_0=\$16000 \end{gathered}[/tex]
By substitution, we have
[tex]\begin{gathered} P(t)=16000\times e^{(0.035\times6)} \\ =19738.84895 \\ \Rightarrow P(t)\approx\$19738.8\text{ \lparen nearest cent\rparen} \end{gathered}[/tex]
Hence, we have
