This question concerns bit strings of length six. These bit strings can be divided up into four types depending on their initial and terminal bit. Thus the types are: 00, 01, 10, 11. How many bit strings of length six must you select before you are sure to have at least 6 that are of the same type?

We have to choose 21 strings to be sure we have chosen at least 6 strings of the same type.

Step-by-step explanation:

Since the string type is determined by the initial and terminal bits as understood from the question, then the value of the bits between the initial and terminal bits is of no concern to us.

Now, to be sure you have atleast 6 of the same type, we select each string five times. By doing this, we have already selected 20 strings because we have 4 strings there. Now if you choose any of the string one more time, we are certain that we must have chosen atleast 6 strings that are the same. This means we have to choose 21 strings to be sure we have chosen at least 6 strings of the same type.

the ratio of white tiles to blue tiles is 48: 16. the ratio in it's simplest form is 3: 1.

step-by-step explanation:

ratio: take the first number and add a colon next to it, then add the second number after that. make sure the ratio order matches up with the written problem!

simplest form: we can simplify the ratio 48 : 16 by dividing both terms by the greatest common factor (gcf).

the gcf of 48 and 16 is 16.

divide both terms by 16:

48 ÷ 16 = 3

16 ÷ 16 = 1

reducing to simplest form:

48 : 16 = 3 : 1