middle half of the data. Find the interquartile
range for the data set: 10, 3, 7, 6, 9, 12, 13.
A 12
© 6
B 7
D 8
X=
The equation for the line containing the vertical right edge of the arch is x = 60.
The line representing the right edge is the line passing through the parabola on the right side at y = 20.
It is given that the left edge of the arch is the y-axis.
The parabola and the vertical base are symmetrical.
We can see that the left edge begins (0, 20). This means that the right edge would also have y = 20.
Observe the given graph. We can see that the parabola passes y = 20 again at the point (60, 20).
This means that the equation of the line representing the right edge is x = 60.
Therefore, we have found the equation for the line containing the vertical right edge of the arch to be x = 60.
Learn more about parabolas here: brainly.com/question/4148030
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Answer:
The answer is x=60.
Since the arch and its base are symmetrical, so that other side of arch should be at (x, 20). Refer to x-axis when y is equal to 20. Thus, x=60.
I just took the test and the answer is correct.
Step-by-step explanation:
the median of the temperatures at Meadows: ______
the interquartile range of the temperatures at Springwood:_______
the interquartile range of the temperatures at Meadows:_______
the difference of the medians as a multiple of their average interquartile range:_______
Hello,
The median of the temperatures at Springwood is 86. (median is at the middle).
the median of the temperatures at Meadows is 73. (median is at the middle).
The (IQR) interquartile range of the temperatures at Meadows is 12. (You have to subtract both of the IQR left and right 80 - 68 = 12).
The (IQR) interquartile range of the temperatures at Springwood is 14. (You have to subtract both of the IQR left and right 91 - 77 = 14).
the difference of the medians as a multiple of their average interquartile range is 79.5 is average median and 13 average IQR.
Answer: The median of the temperatures at Springwood is 86.
The median of the temperatures at Meadows is 73.
The interquartile range of the temperatures at Meadows is 12.
The interquartile range of the temperatures at Springwood is 14.
The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13
i.e. Difference in medians = 1 x (Average interquartile range)
Step-by-step explanation:
In a box - whisker plot , the vertical line in box represents the median value.
The left end of box denotes Lower quartile and the right end of the box Upper quartile.
By considering the given picture,
For Meadows ,
Median = 73
The median of the temperatures at Meadows is 73. (1)
Lower quartile = 68
Upper quartile = 80
Interquartile range = Upper quartile- Lower quartile
Interquartile range=80-68 = 12
The interquartile range of the temperatures at Meadows is 12. (2)
For Springwood,
Median = 86
The median of the temperatures at Springwood is 86. (3)
Lower quartile = 77
Upper quartile = 91
Interquartile range = Upper quartile- Lower quartile
= 91-77 = 14
The interquartile range of the temperatures at Springwood is 14. (4)
Difference in medians = 86-73 = 13 [From (1) and (3)]
Average interquartile range = [From (2) and (4)]
The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13
i.e. Difference in medians = 1 x (Average interquartile range)
Help please :) Im lost and so confused
z=2x(y^2)
z=-6(-4)
z=24