Aquality inspector must verify whether a machine that packages snack foods is working correctly. the inspector will randomly select a sample of packages and weigh the amount of snack food in each. assume that the weights of food in packages filled by this machine have a standard deviation of 0.30 ounce. an estimate of the mean amount of snack food in each package must be reported with 99.6% confidence and a margin of error of no more than 0.12 ounces. what would be the minimum sample size for the number of packages the inspector must select?

52

Step-by-step explanation:

Formula for sample size:-

, where = population standard deviation.

= Two -tailed z-value for (significance level)

E= margin of error.

Given : tex]\sigma=\text{ 0.30 ounce}[/tex]

⇒Significance level for 99.6% confidence level :

By using z-value table ,Two -tailed z-value for :

E= 0.12 ounces.

Minimum sample size will be :-

Hence, the minimum sample size  for the number of packages the inspector must select = 52

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

the correct answer is 5 6 and 7