However, since the exponent has eight bits, this means it has $2^8-1=255$ possible numbers in the interval $-128 \le m \le 127$, our final exponent is $125-127=-2$ resulting in $2^{-2}$. Inserting the sign and the mantissa yields the final number in the decimal representation as $$-2^{-2}\left(1\times 2^0+1\times 2^{-1}+ 1\times 2^{-2}+1\times 2^{-3}+0\times 2^{-4}+1\times 2^{-5}\right)=$$ $$(-0.4765625)_{10}.$$ In this case we have an exact machine representation with 32 bits (actually, we need less than 23 bits for the mantissa).