IEEE754 32-bit single precision format I have a question like this:
Show how the number $-12.75$D is stored in the computer's storage using IEEE754 32-bit single precision format. You are required to show your conversion steps clearly.
My answer is:
$-12.75\mathrm{D}=-1100.11\mathrm{B}$
normalization: $-1.10011\times 2^3$
$E(D)=127+3=130\mathrm{D}$
$E(B)=130\mathrm{D}=1000\ 0010\mathrm{B}$
sign: negative: $1$
$IEEE 754(SP)=1\ 10000010\ 10011....\mathrm{B}$ 
Is it correct?
 A: Your final version is correct.
$\require{enclose}
\def\xD{{}_{10}}
\def\xB{{}_{2}}
$
Given any real number, if its representation in basis $b$ ($2 \le b \le 10$) is given by
a string $???$ consists solely of digits and at most one decimal point, we will use
the notation $???_{b}$ to label it.
Since $$
-12.75\xD = -(2^3 + 2^2 + 0 + 0 + 2^{-1} + 2^{-2}) = -1100.11\xB = -1.10011\xB \times 2^3$$
the sign bit $S$ is $1$, exponent $E$ is $3\xD$ and the mantissa $M$ is 
$1.10011\xB$.
For IEEE754 single precision numbers, we use


*

*$1$ bit for sign, $S = 1 \mapsto \color{red}{\enclose{box}{1}}$

*$8$ bit for exponent but encoded with an offset of $127$. So 
$$E = 3\xD \mapsto 3\xD + 127\xD = 130\xD = (2^7 + 2^1) = 1000 0010\xB
\mapsto \color{green}{\enclose{box}{1000\;0010}}
$$

*$24$ bit for mantissa but the leading bit is implicit and only $23$ bits are stored.
$$M = 1.10011\xB
\mapsto 1\color{blue}{\enclose{box}{100\;1100\;0000\;0000\;0000\;0000}}$$


Under IEEE754, $-12.75\xD$ will be encoded as
$$\color{red}{\enclose{box}{1}}\!-\!\color{green}{\enclose{box}{1000\;0010}}\!-\!\color{blue}{\enclose{box}{100\;1100\;0000\;0000\;0000\;0000}}$$
There are several single precision converter on the web.
The one I used for reference is this. Play with it and it will help you understand how floating points numbers
are encoded in this format.
