Answer
the solution is shown in the table below:
\begin{equation}
\begin{array}{|c|c|c|}\hline & {\text { Binary }} & {\text { Equivalent base } 10 \text { form }} \\ \hline \mathbf{a} & {11.01} & {3\frac{1}{4}} \\ \hline \mathbf{b} & {101.111} & {5\frac{7}{8}} \\ \hline \mathbf{c} & {10.1} & {2\frac{1}{2}} \\ \hline \mathbf{d} & {10.011} & {6\frac{3}{8}} \\ \hline \mathbf{e} & {0.101} & {\frac{5}{8}} \\ \hline\end{array}
\end{equation}
Work Step by Step
a.
Step $1 :$ In the fractional binary notation, the digits to the left of the radix point represent the integer part of the value and are interpreted as in the binary system. $(but)$ The digits to the right represent the fractional part of the value and are interpreted as other bits.
The first position to the right of the radix is assigned the quantity $\frac{1}{2}$ or $({2}^{-1})$, the next position is assigned the quantity $\frac{1}{4}$ or $({2}^{-2})$, and so on.
Step $2 :$ To convert the fractional binary representations to its equivalent base ten representation, we have to multiply every bit of the integer part to the power of $(2)$ from right to left. Start at $(2^0 )$ and multiply every bit of the fractional part to the power of $(\frac{1}{2})$ from left to right.
Step $3 :$ ($2^1$ * 1) + ($2^0$ * 1) $.$ ($2^{-1}$ * 0) + ($2^{-2}$ * 1) = ( 0 + 3) $.$ (0 + $\frac{1}{4}$) = $3\frac{1}{4}$
Therefore, $(11.01)_{2}=(3\frac{1}{4})_{10}$
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to solve [ b, c, d, e ] do the previous steps :
