Convert it! (Problem 4)

$\begin{aligned} 101101_2 & = \left(2^5 + 2^3 + 2^2+2^0\right)_{10} \\ & = (32+8+4+1)_{10} \\ & = \boxed{45}_{10} \end{aligned}$

$101101_2=(1\times 2^0+0\times 2^1+1\times 2^2+1\times 2^3+0\times 2^4+1\times 2^5)_{10}=(1+4+8+32)_{10}=\boxed {(45)_{10}}$ .

Relevant wiki Binary numbers

$101101$ in Base 2 converted to Base 10 is $\boxed{45}$ .