Answer
$$
\begin{array}{|c|c|}\hline & {\text { Solution }} \\ \hline \mathbf{a} & {\frac{5}{8}} \\ \hline \mathbf{b} & {3\frac{1}{4}} \\ \hline \mathbf{c} & { \frac{9}{32} } \\ \hline \mathbf{d} & {-1\frac{1}{2}} \\ \hline \mathbf{e} & { -(\frac{11}{64}) } \\ \hline\end{array}
$$
Work Step by Step
Decode bit pattern
To decode the bit patterns using floating-point format, we designate the high-order bit of the byte as the sign bit. The value stored is positive if the sign bit is 0 and the value is negative if the sign bit is 1.
The Next 3 bits following the sign bit are the exponent field and the remaining 4 bits are represented as mantissa field.
An excess notation system is used for the bit patterns of length
three to represent the value of the corresponding exponent field as
in the below table-
\begin{equation}
\begin{array}{|c|c|}\hline \text { Bit Pattern } & {\text { Value represented }} \\ \hline 111 & {3} \\ \hline 110 & {2} \\ \hline 101 & {1} \\ \hline 100 & {0} \\ \hline 011 & {-1} \\ \hline 010 & {-2} \\ \hline 001 & {-3} \\ \hline 000 & {-4} \\ \hline\end{array}
\end{equation}
$Now$ to Answer $(a)$
we have considered the following floating point notation using one byte of storage:
We designate the high order bit of the byte as the sign bit.
Next, we divide the remaining $7$ bits of the byte into two groups: exponent field using $3$ bit and mantissa field using $4$ bit a.
Analyzing the pattern $(01001010)$ with the above format we get, the sign bit is $(0)$ which represents the number is positive. The exponent is $(100)$ and the mantissa is $(1010)$.
To decode the byte we first extract the mantissa and place a radix point on its left side, obtaining $(.1010)$ Next we extract the exponent component $(100)$ and interpret it as an integer stored using a 3-bit excess method. We get $(0)$ $(100=0 )$.
thus the exponent field represents a zero.
this tells us to move the radix in our solution to the right by 0 bits. Consequently, we get $(.1010)$ which is the binary representation of $( \frac{5}{8} )$
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to solve [ b, c, d, e ] do the previous steps.
