What is the selling price of a $610 couch with a 25% markup?
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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:
Step-by-step explanation:
We'll just work on solving both so you can see what's involved in solving an absolute value equation. Because an absolute value is a distance, we can have that distance being both to the right on the number line of the number in question or to the left. For example, from 2 on the number line, the numbers that are 5 units away are 7 and -3. Using that logic, we will simplify the equation down so we can set up the 2 basic equations needed to solve for x.
If then
What you need to remember here is that you cannot distribute into a set of absolute values like you would a set of parenthesis. The -2 needs to be divided away:
Now we can set up the 2 main equations for this which are
.5x + 1.5 = .5 and .5x + 1.5 = -.5
Knowing that an absolute value will never equal a negative number (because absolute values are distances and distances will NEVER be negative), once we remove the absolute value signs we can in fact state that the expression on the left can be equal to a negative number on the right, like in the second equation above.
Solving the first one:
.5x = -1 so
x = -2
Solving the second one:
.5x = -2 so
x = -4