# Question: i need help with question 3 i found this for...

###### Question details

I need help with question 3, I found this for 1 and 2:

a)-76.678595 = 1001100.10101101101110000 in binary (using the technique in 1)

Now normalize: 1.00110010101101101110000x2^{6}

The mantissa is 00110010101101101110000

The exponent is 6+127=133, which is 10000101

The number is negative so the sign bit is 1

So -76.678595 in single precision format is 1 10000101 00110010101101101110000

In hexadecimal, this is : 0xC2995B70

b)19.459931 = 10011.01110101101 (using the technique in 1)

Now normalize: 1.00110111010110100000000x2^{4}

The mantissa is 001101110101101

The exponent is 4+124=131, which is 10000011

The number is positive so the sign bit is 0

So 19.459931 in single precision format is 0 10000011 00110111010110100000000

In hexadecimal, this is : 0x419BAD00