In a standard game of pool, there are 15 balls labeled 1 through 15. a) In how many ways can the 15 balls be ordered? (5 points) b) In how many ways can 3 of the 15 balls be chosen if order does not matter?

Step-by-step explanation:

If  there are 15 balls labeled 1 through 15 in a standard football game, the order of arrangement of the 15 balls can be done in 15! ways.

15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2

15! = 1.308*10¹² ways

b)  If 3 of the 15 balls are to be chosen if order does not matter, this can be done in 15C3 number of ways. Since we are selecting some balls out of the total number of balls, we will use the concept of combination.

Using the combination formula nCr = n!/(n-r)!r!

15C3 = 15!/(15-3)!3!

15C3 = 15!/12!3!

15C3 = 15*14*13*12!/12!*6

15C3 = 15*14*13/6

15C3 = 455 ways

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;