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sveticcg [70]
3 years ago
14

21,4,9,19,25,16,27,30,33,15,31 Minimum: Quartile1: Quartile2: Quartile3: Maximum:

Mathematics
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

Minimum:4 Quartile 1:15.5 Quartile 2:21 Quartile 3:26 Maximum:33

Step-by-step explanation:

Order out the numbers least to greatest then find the middle number, the middle number is the median or quartile 2, then you find the middle of the first half, that number would be the 1st quartile, then find the number in the middle of the second half, that is your 3rd quartile, the min is the smallest number max is the largest.

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Ugh idkplease help ASAP
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12/23 = 0.5217 = 52%
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What is the rationalizing factor of √3? ​
Vanyuwa [196]

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±  \sqrt{3}  \:  \: ( \sqrt{3}  \:  \: or \:  \:  -  \sqrt{3} )

Step-by-step explanation:

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3 years ago
(Geometry course question) Explain why it is not possible to construct an equilateral triangle that has
Nutka1998 [239]

Answer:

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Step-by-step explanation:

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<h3 /><h3>Location of point C</h3>

With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)

<h3>Equilateral triangle</h3>

The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.

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  \left[\begin{array}{cc}\cos(60^\circ)&\sin(60^\circ)\\-\sin(60^\circ)&\cos(60^\circ)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}x'\\y'\end{array}\right]

Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...

  cos(60°)·x + sin(60°)·y

will be irrational.

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2 years ago
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