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olga_2 [115]
3 years ago
11

A set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. Find the value that is +3 standards dev

iation away from the mean
Mathematics
1 answer:
Sonbull [250]3 years ago
3 0
8.3 × 3 = 24.9

45 + 24.9 = 69.9

Hope this helps!
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If c – 8 is an odd integer, which of the following could be the value of c?
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Answer:

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Step-by-step explanation:

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2 years ago
To factor 9x^2 - 4, you can first rewrite the expression as:
olga2289 [7]

Answer:

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Step-by-step explanation:

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Given the following trigonometric ratio, enumerate the meaning ratio ​
Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}

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\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}

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Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

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\sin \theta = \frac{y}{h} (1)

Cosine

\cos \theta = \frac{x}{h} (2)

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\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x} (3)

Cotangent

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Secant

\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x} (5)

Cosecant

\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y} (6)

Where:

x - Adjacent leg.

y - Opposite leg.

h - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

h = \sqrt{x^{2}+y^{2}}

If y = AC and x = BC, then the trigonometric ratios are presented below:

\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}

\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}

\cot \theta = \frac{BC}{AC}

\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}

\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}

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LekaFEV [45]

Answer:

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