$1.$a i$.$ Convert the decimal number ${333}_{10}$ to its corresponding unsigned binary number.

$[1$ mark$]$

ii$.$ Convert the unsigned $8-$bit binary number ${10111110}_2$ to its corresponding decimal number.

$[1$ mark$]$

iii. Convert the decimal number ${300}_{10}$ to its corresponding hexadecimal number.

$[1$ mark$]$

iv$.$ Convert the following decimal numbers to $8-$bit $2\text{'}$s complement: ${50}_{10},{44}_{10}.$ Then add them, giving the answer also in $8-$bit $2\text{'}$s complement.

$[2$ marks$]$

v$.$ Convert the following decimal numbers to $8-$bit $2\text{'}$s complement: $-{60}_{10},-{55}_{10}.$ Then add them, giving the answer also in $8-$bit $2\text{'}$s complement.

$[2$ marks$]$

$1.$b i$.$ Specify the decimal number $-{128}_2$ in an $8-$bit binary sign and magnitude representation.

$[2$ marks$]$

ii$.$ Specify the hexadecimal $AC{E}_{16}$ in an $8-$bit binary sign and magnitude representation.

$[2$ marks$]$

iii. Convert the decimal number $-\frac{14}{{32}_{10}}$ to the $32-$bit IEEE $754$ floating point and express your answer in hexadecimal. $($Reminder: the $32$ bits are used as follows: Bit $1$: sign of mantissa, bits $2-9$: $8-$bits of exponent in excess $127,$ bits $10-32$: $23$ bits for magnitude of mantissa.$)$

$[6$ marks$]$

$1.$c

i$.$ Using Boolean algebra identities please simplify the following expressions:

a$.,F=AB+$ABC$+$ABCD$+$ABCDE$+$ABCDEF

b$.,F=(A+C)(AD+A{D}^{\text{'}})+AC+C$

$[6$ marks$]$

ii$.$ Produce the desired inverted NAND expression using De Morgan's law on the following sum of products:

a$.,F=A^{\text{'}}{B}^{\text{'}}+$CDE$+A^{\text{'}}+B^{\text{'}}$

$[2$ marks$]$