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IrinaK [193]
3 years ago
14

What is the correct definition of the third quartile of a data set?

Mathematics
2 answers:
algol [13]3 years ago
4 0
We know that 

The third quartile (also called the upper quartile) has 75 percent of the data below it and the top 25 percent of the data above it. <span>The third quartile is the same as the </span>median<span> of the part of the data which is greater than the median. 
</span>
therefore

the answer is the option
<span>C.) The third quartile is the median of the upper half of data in the data set.</span>
shutvik [7]3 years ago
4 0
<h2>Answer:</h2>

Option: C is the correct answer.

C.    The third quartile is the median of the upper half of data in the data set.

<h2>Step-by-step explanation:</h2>

We know that our data is divided into three quartiles:

1)

First quartile or lower quartile--

It is the median of the lower set of the data.

2)

Middle quartile or Median--

Median is the central tendency of the data and it always exist in the middle of the data set.

3)

Third quartile or upper quartile--

It is obtained by finding the median of the upper set of the data.

               Hence, the answer is: Option: C

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larisa [96]

Answer:

15^\circ,165^\circ,15^\circ,165^\circ\\15^\circ,165^\circ,15^\circ,165^\circ

Step-by-step explanation:

The ratio of the same side interior angles of two parallel lines is 33:3

If two parallel lines are cut by a transversal

Sum of the same side interior angles = 180 degrees.

Therefore, the two same-side interior angles are:  

\dfrac{33}{36}\times 180^\circ =165^\circ\\\dfrac{3}{36}\times 180^\circ =15^\circ

We can then say that the measure of all eight angles formed by the parallel lines and transversal starting from the upper left in a clockwise rotation is:

15^\circ,165^\circ,15^\circ,165^\circ, 15^\circ,165^\circ,15^\circ,165^\circ

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3 years ago
Miss Chambers bought 5 pens for $255, if she made a profit of $3 each per pen how much did she sell a pen for?​
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Answer:

$54

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First, we calculate the amount she paid for each pen.

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She paid $51 for each pen.

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