In a box and whisker plot, the Median is represented by the line at the center of the box.
So, from the above figures we can see that the Median for the Class Data is around 82 and the median for the Exam Data is around 87. It is clear that the Median of Exam data is higher than the Median of Class Data.
So, the best statement to describe the medians will be the option C. The class median is lower than the exam median. The option A and B are not correct. The outlier does have not have much of a large impact on the median of the data and neither the spread of the data primarily determines its median. Median only depends on the number of values in the data set.
Therefore, the correct answer is option C
Answer:
https://cdn.kutasoftware.com/Worksheets/Geo/2-Segment%20Addition%20Postulate.pdf Step-by-step explanation:
is this it?
First solve for x in the first equation
x + 7y = 0
x = -7y
since x = -7y we can substitute -7y into the other equation wherever there is an x
2x - 8y = 22
2(-7y) - 8y = 22
-14y - 8y = 22
-22y = 22
y = -1
Now let's substitute -1 back into either of the original equations so we can solve for x
x + 7y = 0
x + 7(-1) = 0
x - 7 = 0
x = 7
So your solution set is
(7, -1)
Slope = -1/5
Coordinate = (1, -4)
We know, y - y1 = m(x - x1)
y + 4 = -1/5 (x - 1)
y + 4 = -1/5x + 1/5
y = -1/5x -19/5
When, x = 1, y = -1/5(1) - 19/5 = -20/5 = -4
x = 2, y = -1/5(2) - 19/5 = -21/5
x = 3, y = -1/5(3) - 19/5 = -22/5
Here, Your Coordinates would be: (0, -4), (1, -21/5), (2, -22/5)
Mark them & draw the lines. Graph is done!
Hope this helps!
