Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.8 ounces. If you took a sample of the 49 bottles of ketchup what would be the approximate 95% confidence interval for a mean number of ounces of ketchup per bottle in the sample?
***ANSWER IS C.24+/-0.029 for apex***

24 ± 0.224

Step-by-step explanation:

Confidence interval = mean ± critical value × standard error

CI = μ ± CV × SE

The critical value is the t score or z score.  Since n > 30, we can approximate using the z score.  For 95% confidence, z = 1.96.

The standard error is:

SE = σ / √n

SE = 0.8 / √49

SE = 0.114

So the confidence interval is:

CI = 24 ± (1.96) (0.114)

CI = 24 ± 0.224

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

to determine the unit cost your would divide the total cost by the number sold.

7 bagels for $4.41       $4.41/7 = $.63 each at fresh side

5 bagels for $3.60         $3.60/5 = $.72 each at value mart