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coldgirl [10]
3 years ago
6

Judy just calculated the standard deviation for a project she is working on. The value for the standard deviation is a small num

ber. What does this represent?
The data is spread far from the mean.

The data is spread close to the mean.

The data is spread equal to the mean.

None of these choices are correct.
Mathematics
1 answer:
Contact [7]3 years ago
7 0

Answer: b) The data is spread close to the mean

Step-by-step explanation: Theoretically, when the standard deviation of a statistical data is large or high, it simply means that the values in such statistical data set are farther away from the mean, On the other hand when the standard deviation is small or low, it simply means that the values of such statistical data are close to the mean of those data on average.

Therefore Judy's low standard deviation for her project simply indicates that her statistical data set is spread close to the mean.

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\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
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sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
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\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
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\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
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\textit{and now, let's rationalize the denominator}
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