Given: ΔABC with Prove: Statement Reason 1. given 2. Addition Property of Equality 3. using common denominators 4. AB = AD + DB
2 answers:
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Answer:
Step-by-step explanation:
Hello!
The variable is X: iron level on human blood.
It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.
According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl
Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl
The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.
Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.
<u>Sample 1</u>
Min= 50.15
Max= 51.99
Mean= 51.61
Without warning 50.22<X[bar]<52.78
Without Action interval 49.58<X[bar]<53.42
<em>This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.</em>
<em><u /></em>
<u>Sample 2 </u>
Min= 50.32
Max= 52.56
Mean= 51.16
Without warning 50.22<X[bar]<52.78
Without Action interval 49.58<X[bar]<53.42
<em>Both the max and min values of the sample are within the working range without warning.</em>
<em />
<u>Sample 3</u>
Min= 50.25
Max= 53.12
Mean= 51.83
Without warning 50.22<X[bar]<52.78
Without Action interval 49.58<X[bar]<53.42
<em>The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.</em>
<em />
<u>Sample 4</u>
Min= 50.05
Max= 53.01
Mean= 51.70
Without warning 50.22<X[bar]<52.78
Without Action interval 49.58<X[bar]<53.42
<em>Both min and max values surpass the warning interval but do not reach action levels</em>.
<u>Sample 5</u>
Min= 50.35
Max= 52.71
Mean= 51.37
Without warning 50.22<X[bar]<52.78
Without Action interval 49.58<X[bar]<53.42
<em>Both min and max values are within working levels without warnings.</em>
<em />
Considering that samples reaching warning levels should be shut down as a precaution, they are classified as:
Shutdown: Sample 1, 3 and 4
Do not shutdown: Sample 2 and 5.
I hope it helps!
Answer:
Diameter of pound = 10 yd
Step-by-step explanation:
Given area of circular pond = 78.5 square yd
To find the diameter of the pond.
Solution:
The pond being circular in shape, the area of pond can be given as:
⇒
where represents radius of the pond.
Thus, we have:
⇒
Using
⇒
Solving for
Dividing both sides by 3.14
⇒
⇒
Taking square root both sides.
⇒
⇒
So, radius = 5 yd, as distances are always positive.
Diameter =
Thus, diameter of pound = 10 yd