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ladessa [460]
3 years ago
9

I need help with both questions on the screen. if you can only answer one thats fine too!

Mathematics
1 answer:
meriva3 years ago
7 0

Answer:

11. b = 30

Step-by-step explanation:

a^{2} + b^{2} = c^{2}

Substitute

16^{2} + b^{2}  = 34^{2}

256 + b^2 = 1156

b^2 = 1156-256\\b^2 = 900

b = \sqrt{900}

b = 30

Your welcome and don't forget to say <u>Thanks</u>

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The box-and-whisker plot below shows the numbers of text messages received in one day by students in the seventh and eighth grad
Leya [2.2K]
<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. Q_1 = 14

Median or second quartile i.e. Q_2 = 18

Third quartile or upper quartile i.e. Q_3 =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. Q_1 = 26

Median or second quartile i.e. Q_2 = 30

Third quartile or upper quartile i.e. Q_3 = 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. Q_3-Q_1

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:

8\times n=12\\\\n=\dfrac{12}{8}\\\\n=\dfrac{3}{2}\\\\n=1.5

Hence, the number is : 1.5

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Answer:

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