 For this example, I'm again going to start with a number in our floating point format, and convert this back into a decimal number. So the first step is to take this number and to put it into the floating point format so we can see what the various fields are. C is 1, 1, 0, 0. 5 is 1, 0, 1. And B is 1, 0, 1, 1. I have 0s for the rest of my precision bits, so they're not going to be terribly important. First I've got a 1, so I have a negative number. We'll have a leading 1, and then the rest of my mantissa. Now I have my exponent. I have 128, which gives me my bias plus 1. And then I have 11. So 11 plus 1 will give me 12. Now I need to convert this back into decimal. I could move my binary point 12 places to the right and figure out what those three bits, the value of those three bits are. Add them up, or I can try something else. I can convert this to 11 times 2 to the 9th. So in this case, I've moved my binary point over three places. Dropped my exponent accordingly, but I still have the exponent. Now, I know that 2 to the 10th is 1,024, so half of that is 512, and I know that this is 11. So really I want to do 11 times 512, which is reasonable to do by hand. 11 times 512 is 5120 plus 512. And this is a negative number because we have one for our sign bit. This floating point number corresponds to negative 5,632 in decimal.