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777dan777 [17]
3 years ago
6

PLZ ANSWER ILL GIVE BRAINILEST2 box plots. The number line goes from 26 to 36. For Pleasantville Middle School, the whiskers ran

ge from 26 to 34, and the box ranges from 29.5 to 33. A line divides the box at 31. For Grandview Middle School, the whiskers range from 29.3 to 35.3, and the box ranges from 30.5 to 33. A line divides the box at 31.8.
PLZ ANSWER ILL GIVE BRAINILEST

Why should you use the mean and mean absolute deviation (MAD) to compare populations with symmetrical distributions?
The mean and MAD will make graphs of the data appear asymmetrical.
The mean and MAD eliminate the influence of outliers.
The mean and MAD can accurately describe the “typical” value in the symmetric data set.
The mean and MAD make the data distribute more unevenly.
Mathematics
2 answers:
Svetach [21]3 years ago
6 0

Answer:

I believe the answer is- The mean and MAD can accurately describe the "typical" value in the symmetric data set.

Step-by-step explanation:

The other answers don't make sense because the mean and MAD are being used for symmetrical distributions and asymmetrical means uneven distributions.

allsm [11]3 years ago
3 0

Answer:

1- mean

2- MAD

Step-by-step explanation:

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uysha [10]

Answer:

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Step-by-step explanation:

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2 years ago
Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE
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Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to 20\ units

Part 3) The area is equal to 25\ units^{2}

Step-by-step explanation:

we have

A(5,0), B(2,4), C(-2,1),D(1,-3)

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance AB

A(5,0),B(2,4)

substitute in the formula

d=\sqrt{(4-0)^{2}+(2-5)^{2}}

d=\sqrt{(4)^{2}+(-3)^{2}}

d=\sqrt{25}

AB=5\ units

Find the distance BC

B(2,4), C(-2,1)

substitute in the formula

d=\sqrt{(1-4)^{2}+(-2-2)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}

BC=5\ units

Find the distance CD

C(-2,1),D(1,-3)

substitute in the formula

d=\sqrt{(-3-1)^{2}+(1+2)^{2}}

d=\sqrt{(-4)^{2}+(3)^{2}}

d=\sqrt{25}

CD=5\ units

Find the distance AD

A(5,0),D(1,-3)

substitute in the formula

d=\sqrt{(-3-0)^{2}+(1-5)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}        

AD=5\ units

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

B(2,4),D(1,-3)

substitute in the formula

d=\sqrt{(-3-4)^{2}+(1-2)^{2}}

d=\sqrt{(-7)^{2}+(-1)^{2}}

BD=\sqrt{50}\ units        

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

BD^{2}=AD^{2}+AB^{2}

substitute

(\sqrt{50})^{2}=5^{2}+5^{2}

50=50 -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

A=b^{2}

we have

b=5\ units

A=5^{2}=25\ units^{2}

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

P=4b

we have

b=5\ units

P=4(5)=20\ units

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ANTONII [103]

Step-by-step explanation:

3 + 3(k + 3) = 6(k - 2) + 9

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12 + 12 - 9 = 6k - 3k

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k = 15/3

k = 5

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