Nov 20'23
Exercise
Nov 20'23
Solution: C
[[math]]
\begin{array}{c}{{s_1=_{1}f_{0} = 0.04 }}\\ {{_{1}f_{1}=0.06=\frac{\left(1+s_{2}\right)^{2}}{\left(1+s_{1}\right)^{2}}-1\Rightarrow s_{2}=\sqrt{(1.06)(1.04995)^{2}})^{1/3}-1=0.04995}}\\ {{\mathrm{}_{1}f_{2}=0.08=\frac{\left(1+s_{2}\right)^{2}}{\left(1+s_{2}\right)^{2}}-1\Rightarrow s_{3}=\mathrm{}\left((1.08\right)(1.04995)^{2}-1=0.05987=6\%.}}\end{array}
[[/math]]