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Alika [10]
2 years ago
14

The data set below has a lower quartile of 13 and an upper quartile of 37.

Mathematics
1 answer:
timama [110]2 years ago
6 0

The correct option regarding the outliers of the data-set is given by:

The greatest value, 78, is the only outlier.

<h3>How to use the quartiles of a data-set to identitfy outliers?</h3>

  • The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
  • The first quartile is the median of the first half of the data-set.
  • The third quartile is the median of the second half of the data-set.
  • The interquartile range is the difference of the third quartile with the first quartile.
  • Measures that are more than 1.5 IQR from Q1 and Q3 are considered outliers.

The IQR for this problem is:

IQR = 37 - 13 = 24.

Hence the bounds for outliers are:

  • Less than 13 - 1.5 x 24 = -23.
  • Greater than 37 + 1.5 x 24 = 73,

The options are:

  • No outliers.
  • Only 1 is an outlier.
  • Only 78 is an outlier.
  • Both 1 and 78 are outliers.

Hence the correct option is that only 78 is an outlier.

More can be learned about outliers of a data-set at brainly.com/question/17083142

#SPJ1

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Dahasolnce [82]
I think you would use the Pythagorean Theorem to solve this, as a square cut across diagonally creates two isocele triangles. Since the longest side is 20 m, this value would be imputed into c^2.

a^2 + b^2 = c^2
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7 0
3 years ago
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Answer:

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Step-by-step explanation:

Given

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Required

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The first thing to do is to check through the set to count the number of occurrence of -2

Looking through Set = {−6,−3,−5,−1}, we have that the occurrence of -2 is 0

This implies that -2 is not an element of the given set and this is represented by the symbol ∉

Hence:

−2​ ∉ {−6,−3,−5,−1}

7 0
3 years ago
How many cubic centimeters of water can this paper cone cup hold?
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Answer:

150.72 cubic centimeter

Step-by-step explanation:

Given:-

Height of cone (h) = 9cm

Diameter of cone(d)=8cm

radius of cone(r)=\frac{d}{2}=\frac{8}{2} =4cm

To calculate= Volume of cone(V_{cone})

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4 0
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