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PSYCHO15rus [73]

2 years ago

15

If the variance of the "number of daily parking tickets issued is 100, the standard deviation is defined as the _____________. A

. Absolute value of the variance of the "number of daily parking tickets" B. "Number of daily parking tickets" squared. C. "Number of daily parking tickets" D. Square root of the variance of the "number of daily parking tickets"

1 answer:

Alex17521 [72]2 years ago

8 0

Answer:

Square root of the variance of the "number of daily parking tickets"

Step-by-step explanation:

The Standard Deviation is a measure of how spread out numbers are.

Its symbol is σ (the Greek letter sigma)

The formula is easy: it is the square root of the Variance.

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Answer:

54%

Step-by-step explanation:

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2 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

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Answer:

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Step-by-step explanation:

Given

$u =$

$v =$

Required:

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For a vector

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$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

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For a vector

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$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

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$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

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So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

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$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

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~~~~~~~~~~~~~~

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Answer:

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