Given a principal of P and a depreciation rate of r %,
after n years the new amount A is given by
[tex]A=P(1-r^{})^n[/tex]In this case,
[tex]A=\text{ \$2500 r = 12\% n = 4years}[/tex]Then, we have
[tex]\begin{gathered} 2500=P(1-0.12^4) \\ \text{ Which implies that } \\ P=\frac{2500}{(1-0.12)^4}=\frac{2500}{0.88^4}=\frac{2500}{0.5997}=\text{ \$4169} \end{gathered}[/tex]Hence the cost brand new is $4169