## Intermediate Algebra (6th Edition)

$\frac{1}{16}$
We are given the expression $4^{-2}$. In general, $a^{-n}=\frac{1}{a^{n}}$, where a is a nonzero real number and n is a positive integer. Therefore, $4^{-2}=\frac{1}{4^{2}}=\frac{1}{4\times4}=\frac{1}{16}$