Amanager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium shells. past data indicate that if the machine is functioning properly, the length of the shells produced by this machine can be modeled as being normally distributed with a mean of 118 centimeters and a standard deviation of 6.3 centimeters. suppose 10 shells produced by this machine are randomly selected. what is the probability that the average length of these 10 shells will be between 116 and 120 centimeters when the machine is operating "properly? "

0.6826

Step-by-step explanation:

To solve this, we find the z scores for both sample means.  We then us a z table to find the area under the curve to the left of (probability less than) each z score, and subtract them to find the area between them.

The formula we use, since we are using sample means, is

Our x-bar will be 116 in the first z-score and 120 in the second; our mean, μ, is 118; our standard deviation, σ, is 6.3; and our sample size, n, is 10:

Using a z table, we see that the area under the curve to the left of -1.00 is 0.1587.  The area under the curve to the left of 1.00 is 0.8413.  This makes the area between them

0.8413-0.1587 = 0.6826.

1/8 3/16 3/4

step-by-step explanation:

you put them as all the same denominator and multiply

60.9774

step-by-step explanation:

because when you add 47.6 plus 7.32 equals 54.92

and when you add .0574 + 6= 6.0574

and when you add 54.92 and 6.0574 you get the answer of 60.9774.

plzz mark brainlest last one got deleted. cause i didn't explain it but i did but i understand what the mod meant