If there were 80 E. coli bacteria in the sausage roll at 10am and bacteria divide into 2 every 20 minutes, how many bacteria were in the sausage...

We are told that the initial E. coli population was 80 bacteria (at 10am) and the population doubles every 20 minutes. We are asked to find out how many bacteria are present at 2pm. 

Four hours have passed between 10am and 2pm. This is equal to 12 twenty minute periods. (1 hr = 3 twenty minute periods). Thus, the bacterial population has doubled 12 times over the course of the four hours.

To calculate the size of the population when the roll was eaten, we multiply the starting population by the number 2 raised to the number of periods, as follows:

`Pf=(Po)*2^(12)=80*2^(12)=327680`

Where Pf is the final population and Po is the starting population. In conclusion, 327,680 E. coli bacteria were present in the roll at the time it was eaten.