Chapter 11 - Polynomial and Rational Functions - 11.1 Power Functions and Proportionality - Exercises and Problems for Section 11.1 - Exercises and Problems - Page 439: 4

Yes, $R(t)=t^{-\frac{1}{2}}$

Since $\sqrt{ab}=\sqrt{a} \cdot \sqrt{b}$, $$R(t)=\frac{4}{\sqrt{16} \cdot \sqrt{t}}=\frac{4}{4\sqrt{t}}=\frac{1}{\sqrt{t}}$$ Using the fact that $\sqrt{t}=t^\frac{1}{2}$ and that $\frac{1}{a^n}=a^{-n}$, the function becomes $R(t)=t^{-\frac{1}{2}}$, a power function with $k=1$ and $p=-\frac{1}{2}$.