First 6, then 11, then 13.5, then (honestly I don't know the rest). '__'
The easiest thing to find in a table showing a linear relationship is the slope. If the x is increasing by one each and every time (or cell because we are talking about tables), then the difference between each y value is the slope. Also, if you are lucky enough to have that sort of table, you can easily find the y-intercept by looking at the y value that is next to the x value of 0. Otherwise, work backward to find the y-intercept. For every x value you go down to try and get to 0, minus the y values also by the slope.
Answer: 
Factor
using the AC method.
Answer: There is 0.014 proportion of connectors fail during the warranty period.
Step-by-step explanation:
Let A be the event that connector fails.
Let B be the event that connector wet.
P(B) = 10%
P(B') = 90%
P(A|B) = 0.05
P(A|B') = 0.01
So, we need to find the proportion of connectors fail during the warranty period is given by

Hence, there is 0.014 proportion of connectors fail during the warranty period.
Answer:
See explaination
Step-by-step explanation:
Refer to attached file for table used in solving mean.
The mean of range is
\bar{R}=\frac{13.3}{20}=0.665
The mean of all six means:
\bar{\bar{x}}=\frac{1907.96}{20}=95.398
(a)
Here sungroup size is 5:
Range chart:
From constant table we have
D_{4}=2.114
So upper control limit:
UCL_{R}=D_{4}\cdot \bar{R}=2.114\cdot 0.665=1.40581
Lower control limit:
LCL_{R}=0.0000
Central limit: \bar{R}=0.665
Since all the range points are with in control limits so this chart shows that process is under control.
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X-bar chart:
From constant table we have
A_{2}=0.577
So upper control limit:
UCL_{\bar{x}}=\bar{\bar{x}}+A_{2}\cdot \bar{R}=95.398+0.577\cdot 0.665=95.78
Lower control limit:
LCL_{\bar{x}}=\bar{\bar{x}}-A_{2}\cdot \bar{R}=95.398-0.577\cdot 0.665=95.01
Central limit: \bar{\bar{x}}=95.398
Sample number 94.82 is not in teh limits of x-bar chart so it seems that process is not in control