Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set - Page 408: 19

$16$

Using $x^{-a}=\dfrac{1}{x^a}$ the given expression, $ \left( \dfrac{1}{4} \right)^{-2} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{1}{\left( \dfrac{1}{4} \right)^{2}} \\\\= \dfrac{1}{\left( \dfrac{1}{16} \right)} \\\\= 1\div\dfrac{1}{16} \\\\= 1\cdot\dfrac{16}{1} \\\\= 16 .\end{array}