Answer
$\frac{1}{16}$
Work Step by Step
We are given the expression $(2^{-2})^{2}$.
To simplify, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers).
$(2^{-2})^{2}=2^{-2\times2}=2^{-4}$
To simplify this into a positive exponent, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer).
Therefore, $2^{-4}=\frac{1}{2^{4}}=\frac{1}{2\times2\times2\times2}=\frac{1}{16}$
