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Alex17521 [72]
2 years ago
12

Which statement is true about the box plots? Both the ranges and the interquartile ranges for the data sets are the same. Neithe

r the ranges nor the interquartile ranges for the data sets are the same. The interquartile ranges for the box plots are the same

Mathematics
2 answers:
kati45 [8]2 years ago
7 0

Answer:

Neither the ranges nor the interquartile ranges for the data sets are the same.

Step-by-step explanation:

In a visual display, the boxplot presents five sample statistics: the minimum, the lower quartile, the median, the upper quartile and the maximum, and the box length gives an indication of the sample variability and the line across the box shows where the sample is centred, with an end at each quartile. The length of the box is thus the interquartile range of the sample and, whether the sample is symmetric or skewed, either to the right or left, the "shape" of the sample, and by implication, the shape of the population from which it was drawn, considering appropriate analyses of the data.

Semenov [28]2 years ago
5 0

The question is incomplete! The complete question along with answer and explanation is provided below.  

Question:  

Which statement is true about the box plots? (attached in the image)

A Both the ranges and the interquartile ranges for the data sets are the same.

B. Neither the ranges nor the interquartile ranges for the data sets are the same.

C. The interquartile ranges for the box plots are the same, but their ranges are different.

D. The ranges for the box plots are the same, but their interquartile ranges are different.

Answer:

D. The range of both box-pots is same (9) but the interquartile range of box-plots is different (6 and 5)

Step-by-step explanation:

To answer this question, first we have to understand what a box plot is!

A box plot is a type of graph which shows 5 statistical characteristics of a data set.

1. Maximum and 2. Minimum values of data

3. Upper Interquartile and 4. Lower interquartile of data

5. Median of the data

Now lets analyze the attached box-plot so that we can conclude what is true about them and what is not!

We have two box-plots for two teacher's classes Marc and Sue and they show the number of incorrect questions in exam.

For Sue's class: (on the bottom)

As you can see the maximum and minimum values are

Maximum = 12 and Minimum = 3

So the Range becomes = 12 - 3 = 9

The upper quartile Q3 is 10 and lower quartile Q1 is 5

So the Interquartile Range becomes = 10 - 5 = 5

This Interquartile Range represents the 25 to 75 percentile of the data

There is little vertical line skewed to the right represents the Median = 9

So to summarize Sue's class

Range = 9

Interquartile Range = 5

Median = 9

For Marc's class: (on the top)

As you can see the maximum and minimum values are

Maximum = 10 and Minimum = 1

So the Range becomes = 10 - 1 = 9

The upper quartile Q3 is 9 and lower quartile Q1 is 3

So the Interquartile Range becomes = 9 - 3 = 6

This Interquartile Range represents the 25 to 75 percentile of the data

There is little vertical line skewed to the left represents the Median = 5

So to summarize Marc's class

Range = 9

Interquartile Range = 6

Median = 5

Conclusion:

The range of both box-pots is same (9) but the interquartile range of box-plots is different (6 and 5).

Therefore, we can confidently conclude that option D is the correct answer.

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Katyanochek1 [597]

Answer:

x = 2 ± (i)√5  (Answer b)

Step-by-step explanation:

x²-4x+9=0 can be solved in a variety of ways; the first two that come to mind that are also appropriate are (1) completing the square and (2) using the quadratic formula.

Completing the square is fast here:

Rewrite x²-4x+9=0 as   x²-4x                    +9=0

Identify the coefficient of the x term:  it is -4

Take half of that, obtaining -2

Square this result, obtaining 4

Add 4 to   x²-4x                    +9=0, in the blank space in the middle, and then subtract 4:      x²-4x +4          -4         +9=0

Rewrite           x²-4x +4 as the square of a binomial:

                          (x - 2)² - 4 + 9  = 0  →    (x - 2)²  = -5

Take the square root of both sides:       x - 2  = ±√(-5)  =  ± (i)√5

Then x = 2 ± (i)√5

4 0
2 years ago
Відрізки ас і вд перетинаються в точці о в поділяються нею навпіл. XC= 24 чому дорвінює відстань DУ
Eduardwww [97]

<em>ОС = 7,2 см.</em>

<em>ОС = 7,2 см.Объяснение:</em>

<em>ОС = 7,2 см.Объяснение:Рисунок прилагается.</em>

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7 0
2 years ago
The length of the base edge of a pyramid with a regular hexagon base is represented as x. The height of the pyramid is 3 times l
sergij07 [2.7K]

Answer:

(a)

h=3x

(b)

A=\frac{\sqrt{3} }{4} x^2

(c)

A=\frac{3\sqrt{3} }{2} x^2

(d)

V=\frac{3\sqrt{3} }{2} x^3 units^3

Step-by-step explanation:

We are given a regular hexagon pyramid

Since, it is regular hexagon

so, value of edge of all sides must be same

The length of the base edge of a pyramid with a regular hexagon base is represented as x

so, edge of base =x

b=x

Let's assume each blank spaces as a , b , c, d

we will find value for each spaces

(a)

The height of the pyramid is 3 times longer than the base edge

so, height =3*edge of base

height=3x

h=3x

(b)

Since, it is in units^2

so, it is given to find area

we know that

area of equilateral triangle is

=\frac{\sqrt{3} }{4} b^2

h=3x

b=x

now, we can plug values

A=\frac{\sqrt{3} }{4} x^2

(c)

we know that

there are six such triangles in the base of hexagon

So,

Area of base of hexagon = 6* (area of triangle)

Area of base of hexagon is

=6\times \frac{\sqrt{3} }{4} x^2

=\frac{3\sqrt{3} }{2} x^2

(d)

Volume=(1/3)* (Area of hexagon)*(height of pyramid)

now, we can plug values

Volume is

=\frac{1}{3}\times\frac{3\sqrt{3} }{2} x^2\times (3x)

V=\frac{3\sqrt{3} }{2} x^3 units^3


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