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zloy xaker [14]
2 years ago
12

Two data sets contain an equal number of values. The double box and whisker plot represents the values in the data set. Compare

the data sets using measures of center and variation.

Mathematics
2 answers:
Ipatiy [6.2K]2 years ago
5 0

Answer:

ive seen this before

Step-by-step explanation:

erma4kov [3.2K]2 years ago
4 0

Answer with explanation:

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

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Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

But

(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

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2 years ago
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