## Algebra 1

$\frac{1}{8}$
The given expression : $\frac{(2^{4})^{3}}{2^{15}}=$ $\frac{2^{12}}{2^{15}}$ , (since $(a^{m})^{n}=$ $a^{mn}$) This can be written as : $2^{12}$ X $2^{-15}$ (since $a^{-1}=$ $\frac{1}{a}$) Thus, this becomes : $2^{-3}=$ $\frac{1}{8}$