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laila [671]
3 years ago
15

Which of the following sets of possible side lengths forms a right triangle?

Mathematics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

Which of the following sets of possible side lengths forms a right triangle?

<h3><em><u>11, 60, and 61</u></em><em><u>✓</u></em></h3>

6, 12, and 13

9, 40, and 45

12, 35, and 38

Step-by-step explanation:

\sqrt{ {11}^{2} +  {60}^{2}  }   \\ =  \sqrt{121 + 3600}  \\  =  \sqrt{3721}  \\  =  \sqrt{ {61}^{2} }  \\ =  61

<h3>11, 60 and 61 is the right answer.</h3>
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In variables estimation sampling, the sample standard deviation is used to calculate the?
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The sample standard deviation is used to calculate the determine the spread of estimates for a set of observations (i.e., a data set) from the mean (average or expected value).

<h3>What is sample standard deviation?</h3>

The spread of a data distribution is measured by standard deviation. The average distance between each data point and the mean is measured.

The sample standard deviation (s) is a measurement of the variation from the expected values and is equal to the sample variance's square root.

where

s = sample standard deviation

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xi= the observed values of a sample item

\overline {x}= the mean value of the observations

Learn more about simple standard deviation, refer:

brainly.com/question/26941429

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2 years ago
10 points! ILL GIVE BRAINLIST! ( if explanation is really good or well/okay)
mel-nik [20]

Answer:

1) 51.12

2)1

3)b

Step-by-step explanation:

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3 years ago
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jasenka [17]
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Answer please thank you . Make her you like !
mariarad [96]

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The answer would be -4.36.

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4 years ago
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How to solve this :') please help
blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\

Use \cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1

so

\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}

7 0
2 years ago
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