A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a random sample of 25 ravioli and measure the weight of cheese filling. They find a sample mean weight of 15 grams with a standard deviation of 1.5 grams. What is the margin of error at 90% confidence?

The margin of error at 90% confidence is 0.5133 gram.

Step-by-step explanation:

Margin of error (E) = (critical value × sample standard deviation) ÷ sqrt(n)

sample standard deviation = 1.5 grams

confidence level = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

n (sample size) = 25

degree of freedom = n - 1 = 25 - 1 = 24

critical value (t) corresponding to 24 degrees of freedom and 10% significance level is 1.711

E = (1.711×1.5) ÷ sqrt(25) = 2.5665 ÷ 5 = 0.5133 gram

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90