Solution:
Using the Maclaurin series;
[tex]e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...[/tex]Thus;
[tex]\begin{gathered} x=0.2; \\ \\ e^{0.2}=1+(0.2)+\frac{(0.2)^2}{2!}+\frac{(0.2)^3}{3!}+\frac{(0.2)^4}{4!} \\ \\ e^{0.2}\approx1.22 \\ \end{gathered}[/tex]ANSWER: 1.22