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Answer: The required width of the park is 3 miles.
Step-by-step explanation: Given that a 5 mile jogging path diagonal divides a rectangular park in half and the park is 4 miles long.
We are to find the width of the park.
As shown in the attached figure below, the diagonal BD divides the rectangle ABCD into two equal parts,
where length DA = 5 miles, BD = 4 miles and AB = width = ?
Since each angle of a rectangle is right-angle, so triangle ABD will be a right-angled triangle.
Using Pythagoras theorem in triangle ABD, we have
Thus, the required width of the park is 3 miles.
<h3><u>Answer with explanation:</u></h3>
From the box and whisker plot of seventh grade we have:
The minimum value =6
First quartile or lower quartile i.e. = 14
Median or second quartile i.e. = 18
Third quartile or upper quartile i.e. =22
and maximum value = 26
From the box and whisker plot of eighth grade we have:
The minimum value =22
First quartile or lower quartile i.e. = 26
Median or second quartile i.e. = 30
Third quartile or upper quartile i.e. = 34
and maximum value = 38
a)
The overlap of the two sets of data is as follows.
- The upper quartile or third quartile of seventh grade is same as the minimum value of the data of eighth grade.
- And the maximum value of seventh grade is same as the lower quartile of eighth grade.
b)
IQR is calculated as the difference of the Upper quartile and the lower quartile
i.e.
so, IQR of seventh grade is:
22-14=8
IQR of seventh grade=8
IQR of eighth grade is:
34-26=8
Hence, IQR of eighth grade=8
c)
The difference of the median of the two data sets is:
30-18=12
Hence, the difference of median is: 12
d)
As the IQR of both the sets is same i.e. 8.
Hence, the number that must be multiplied by IQR so that it is equal to the difference between the medians of the two sets is:
Hence, the number is : 1.5