Answer:
I think it is 1
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
This problem is analogous to the extraction of 6 elements from a total of 10 elements. It's the same if they are marbles, chips, or in this case, people, as here we don't care about the order of the selection as we only are drawing a sample.
Thus, the problem implies solving the amount of possible combinations of 10 people if we take by 6. There is a formula for this and is:
10 C 6 = 10!(6!4!)
If we operate, knowing that for any number x, x!=x*(x-1)*(x-2)*...*1
10 C 6 = 10!(6!4!) = 10*9*8*7*6*...*1 / [(6*5*...*1) * (4*3*2*1)]
10 C 6 = 10*9*8*7*6! / [(6*5*...*1) * (4*3*2*1)]
We have a 6! multiplying and another dividing, so they get eliminated, and as 4*2=8 and 9=3*3
10 C 6 = 10*9*8*7*/ [(4*3*2*1)] = 10*3*3*8*7*/ [(8*3*1)]
We can eliminate the 8s and one of the 3s on the numerator with the one on the denominator:
10 C 6 = 10*3*7*/1 = 210/1= 210
So, option C
1/2 pounds,
assuming that each friend takes an equal part of the 2 1/2 pounds, then 2 1/2 divided by 5 would get you 1/2 pound
Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if 
and 
then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.
side note: q = 1-p is the complement of probability p
Answer:
Step-by-step explanation:
i need indivdual prices to make an equation for the proper answer.
