Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 62

Answer

$\frac{4}{27}$

Work Step by Step

According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$. Therefore, $\frac{3^{-3}}{2^{-2}}=\frac{1}{\frac{3^{3}}{2^{2}}}=1\div\frac{3^{3}}{2^{2}}=1\times\frac{2^{2}}{3^{3}}=1\times\frac{4}{27}=\frac{4}{27}$.
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