Let's break down the expression $\frac{8^{-1} \times 5^3}{2^4}$.

1. Evaluate $8^{-1}$:
   $8^{-1} = \frac{1}{8}$, since raising to the power of $-1$ means taking the reciprocal.

2. Evaluate $5^3$:
   $5^3 = 5 \times 5 \times 5 = 125$.

3. Evaluate $2^4$:
   $2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Now, substitute these values back into the expression:

$\frac{\frac{1}{8} \times 125}{16} = \frac{125}{8 \times 16} = \frac{125}{128}$

So, the final result is:

$\frac{125}{128}$

Would you like a detailed explanation of any step?

Here are 5 related questions:

1. What is $8^{-2} \times 5^2$?
2. How do you simplify $\frac{4^{-1} \times 7^2}{3^3}$?
3. What is the result of $\left( \frac{6^2}{3^2} \right)^2$?
4. How do you evaluate $2^{-3} \times 3^4$?
5. Can you simplify $\frac{9^2}{3^4}$?

Tip: Negative exponents flip the base, turning $a^{-n}$ into $\frac{1}{a^n}$.