A fair 6-sided die is rolled.  If I roll $n$, then I win $n^2$ dollars.  What is the expected value of my win?  Express your answer as a dollar value rounded to the nearest cent.
Each number from 1 to 6 has probability $\dfrac16$ of being rolled, so the expected value is \begin{align*}
E &= \left(\dfrac{1}{6}\times \$1^2\right) + \left(\dfrac{1}{6}\times \$2^2\right) + \cdots + \left(\dfrac{1}{6} \times \$6^2\right) \\
&= \dfrac{1}{6}(\$1 + \$4 + \$9 + \$16 + \$25 + \$36) \\
&= \$\dfrac{91}{6} \\
& \approx \boxed{\$15.17}.
\end{align*}