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Bezzdna [24]
3 years ago
11

The box plots show the weights, in pounds, of the dogs in two different animal shelters.

Mathematics
2 answers:
Dmitriy789 [7]3 years ago
7 0

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

The median weight for shelter A is greater than that for shelter B.

The median of A = 21 and the median of B = 18  true

The median weight for shelter B is greater than that for shelter A.

The median of A = 21 and the median of B = 18   false

The data for shelter A are a symmetric data set.

False, looking at the box it is not symmetric

The data for shelter B are a symmetric data set.

true, looking at the box it is  symmetric

The interquartile range of shelter A is greater than the interquartile range of shelter B.

IQR = 28 - 17 = 11 for A

IQR for B = 20 -16 = 4  True

ANTONII [103]3 years ago
7 0

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

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A 20° sector in a circle has an area of 684 yd2.
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3 years ago
Why is (x-3) NOT a factor of x3+2x2-5x-6
ivolga24 [154]

As the expression is not equal to zero on x=3, x-3 is not a factor of given expression

Step-by-step explanation:

Given expression is:

x^3+2x^2-5x-6

We have to check if x-3 is a factor of given expression

In order for x-3 to be a factor of expression, we have to put x-3 = 0 => x=3 in expression.

If the expression is zero at x=3 then x-3 is a factor of expression otherwise not.

So putting x=333 in expression

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As the expression is not equal to zero on x=3, x-3 is not a factor of given expression

Keywords: Polynomials, expressions

Learn more about polynomials at:

  • brainly.com/question/4228574
  • brainly.com/question/4279146

#LearnwithBrainly

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