Convert this number from one base to another: $(24\mathrm{.}14{)}_8$ to Hexadecimal $()16$

a$.(24\mathrm{.}3{)}_8$

c$.(14\mathrm{.}14{)}_{16}$

b$.(14\mathrm{.}3{)}_{16}$

d$.$ None

Convert this number from one base to another: $(184)10$ to $$CD

a$.(111001111000)$ BCD

c$.(000110000100)BCD$

b$.(10111001)BCD$

d$.(184)BCD$

The required number of registers to represent $(10{)}_{10}$ in binary

a$.,2-$registers $(2-$bits$)$

c$.4-$registers $(4-$bits$)$

b$.3-$registers $(3-$bits$)$

d$.8-$registers $(8-$bits$)$

The $10\text{'}$s complement of $(726400{)}_{10}$

a$.384600$

c$.953300$

b$.753200$

d$.273600$

The result of subtracting $(1100010{)}_2$ from $(1011001{)}_2$ using $2\text{'}$s complement is

a$.0111001$

c$.1000011$

b$.1100101$

d$.0110110$

The sign magnitude representation of $-7$ is

a$.00000111$

c$.11111111$

b$.10000111$

d$.10010111$

If you are given a word of size $n$ bits, the range of $2\text{'}$s complement of binary numbers is

a$.-2^{n+1}$ to $+2^{n+1}$

c$.-2^{n-1}$ to $+2^{n+1}$

b$.,-2^{n-1}$ to $+2^{n-1}$

d$.-2^{n-1}$ to $+2^{n-1}-1$

What is the $BCD$ decimal number $29\mathrm{.}25$

a$.,11101\mathrm{.}010$

c$.00101001\mathrm{.}010$

b$.,11101\mathrm{.}100$

d$.00101001\mathrm{.}00100101$

The representation of the binary number $(110\mathrm{.}0111{)}_2$ in Octal is

a$.(6\mathrm{.}34{)}_8$

c$.,(6\mathrm{.}07{)}_8$

b$.(6\mathrm{.}7{)}_8$

d$.(6\mathrm{.}37{)}_8$