Answer:
We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .
Step-by-step explanation:
Given -
The sample size is large then we can use central limit theorem
n = 50 ,
Standard deviation
= 7.1
Mean
= 110
1 - confidence interval = 1 - .98 = .02
= 2.33
98% confidence interval for the mean caffeine content for cups dispensed by the machine = 
= 
= 
First we take + sign
= 112.34
now we take - sign
= 107.66
We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .
Answer:

<h3><u>a=19.5</u> is the right answer.</h3>
The answer is 2/3 x 18= 12.
Answer:
12
Step-by-step explanation:
Since M is a midpoint of LN, we can see 3x equal to x+8. Remove 1 x to get 2x=8. Solve for x to get x=4. <em>Note that this is not the end to your problem. They asked for the length of MN, not the value of x.</em><em> </em>Plugging x into MN give you 4+8, which is 12.
(If you need to show work, here it is)
3x=x+8
2x=8
x=4
MN=x+8
MN=4+8
MN=12