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.

stew reads at the fastest rate of 265 words per minute.

step-by-step explanation:

it's a given that felicia reads at 260 word per minute and on connie's graph we can see that she reads 200 words per minute. since stew's table tells us that he read 795 in 3 minutes, we can divide 795 by 3 to find out how many words he reads per minute.

795/3 = 265

stew reads 265 words per minute - he reads at the fastest rate since felecia reads at 260 words per minute and connie reads at 200 words per minute.

stew (265) > felecia (260) > connie (200)

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23