### Instruction: Find $\left(\frac{1}{2}\right)^{8} \cdot \left(\frac{3}{4}\right)^{-3}$.### Response: To simplify this expression, I can use the exponent rules that say $(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$.
So, $\left(\frac{1}{2}\right)^{8} = \frac{1}{2^8}$ and $\left(\frac{3}{4}\right)^{-3} = \frac{1}{\left(\frac{3}{4}\right)^3} = \frac{4^3}{3^3}$.
Now, I can multiply the fractions by multiplying the numerators and the denominators: $\frac{1}{2^8} \cdot \frac{4^3}{3^3} = \frac{4^3}{2^8 \cdot 3^3}$.
I notice that $4^3 = 2^6$, so I can simplify the fraction by canceling out the common factors of 2: $\frac{2^6}{2^8 \cdot 3^3} = \frac{1}{2^2 \cdot 3^3} = \frac{1}{4 \cdot 27}$.
Finally, I can write the fraction as a decimal by dividing the numerator by the denominator: $\frac{1}{4 \cdot 27} = 0.009259259 \dots$.