A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 442 gram setting. It is believed that the machine is underfilling the bags. A 44 bag sample had a mean of 438 grams. Assume the population variance is known to be 576. A level of significance of 0.1 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

p value is 0.1343

Step-by-step explanation:

Null: u>= 442

Alternative: u < 442

Using the formula for z score:

(x - u)/sd/√n

Where x is 438, u = 442 sd can be determined from the variance = √variance =√576 = 24 and n = 44

z score = 438-442 / (24/√44)

z score = -4/(24/6.6332)

z = -4/3.6182

z =-1.1055

Now let's find the p value at 0.1 significance level using a z score of -1.1055, using a p value calculator, p value is 0.1343 which greatest than 0.1 meaning the day is not sufficient enough to conclude that the machine is underfilling the bags.

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week

the answer is  

a.

4,624π mm2

step-by-step explanation: