The percentage of data that is roughly greater than 66, as displayed in the box plot, is 100%.
<h3>How to Determine a Percentage of a Data Represented in a Box Plot?</h3>
In a box plot, we have the following displayed five-number summary which tells what percentage of the data distribution for each part of the data distribution:
Upper quartile (Q3): This is the value at where the box in the box plot ends at the edge of the box. From this point to the left, all data values that fall within the bracket make up 75% of the data.
Lower quartile (Q3): This is the value at where the box in the box plot starts at the edge of the box. From this point to the left, all data values that fall within the bracket make up 25% of the data.
Median: this is the middle value at the point where the line divides the box and data below this point make up 50% of the data.
The other five-number summary are the maximum and the minimum values that are represented by the whiskers.
On the box plot given, 66 is at the extreme whisker at the left. This means that the percentage of data that is roughly greater than 66 is 100%.
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<span>Asking a student if they have violated the honor code is not the proper question to ask, for a couple of reasons. First is that maybe they don't even know what the honor code is, and second maybe the students aren't being honest in fear of the reprecussions.</span>
The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24
<h3>How can the polynomial with fractions be simplified to find<em> </em><em>x</em>?</h3>
The given equation is presented as follows;
Factoring the common denominator, we have;
Simplifying the numerator of the right hand side using a graphing calculator, we get;
By expanding and collecting, the terms of the numerator gives;
-(x³ - 48•x + 651•x - 2548)
Given that the terms of the numerator have several factors in common, we get;
-(x³ - 48•x + 651•x - 2548) = -(x-7)•(x-28)•(x-13)
Which gives;
Which gives;
x - 4 = 20
Therefore;
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Ummm, we can't choose one of the problems if we can't see any of the problems...
