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).