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3 years ago

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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3 years ago

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Answer:

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Step-by-step explanation:

Let's say that there are s student tickets and n nonstudent tickets.

s + n = 669

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Multiply -5 by the entire first equation to get...

-5s - 5n = -3345

Combine the two equations.

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Now, put the number of nonstudent tickets sold back into the first equation.

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Hope this helps!

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