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3 years ago

Please help! Question above (+10 points)

Mathematics

1 answer:

cricket20 [7]3 years ago

3 0

There are three answers and they are: choice 2, choice 3, choice 5

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Further Explanation:

Choice 1 is false because the intersection of the altitudes of a triangle leads to the orthocenter.

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Choice 2 is true because the three medians of any triangle always intersect at the centroid. A median is a line that goes from one vertex to the midpoint of the opposite side. In this case, we go from point E to the midpoint of side CD. The midpoint of CD is found by bisecting segment CD (see choice 5)

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Choice 3 is true. This is effectively the same as choice 5 below. The "perpendicular" aspect does not matter.

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Choice 4 is false. Following the steps mentioned here will create an altitude line (see choice 1)

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Choice 5 is true. To bisect something is to cut it in half. Let's say that point F is the midpoint of line segment CD. This means that line segment EF is one of the three medians of triangle CDE.

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edit update: I changed choice 3 from false to true

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Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

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As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

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So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

For First Half:

98, 100, 101, 102, 108, 109, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

For Second Half:

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Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

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