Please help!!

The DVD disk company makes DVD's to be recorded on at are 120 mm wide. The size of the manufactured disks have a standard deviation of 1 mm. The
company manufactures 1000 DVD's an hour.

A)What % of the DVD's would you expect to be greater than 120 mm wide?

B)In one hour how many DVD's would you expect to be between 119 mm
and 121 mm?
DVD's are rejected if they are more than 122 mm wide or less than 118 mm wide? How many DVD's would you expect to be rejected one hour?

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

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;

answer: -0.5 would be the most likley outcome

step-by-step explanation: