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masya89 [10]
3 years ago
14

Which question(s) could be answered by collecting data that has variability? Click on each statistical question. To change an an

swer, click again. How many people are in Mr. Moore's geometry class? What proportion of the voters want a new city park? What was the temperature in the school lunchroom at noon today? How many people own snow blowers in Boston? On average, how old are the cats that live in Kyle's neighborhood?
Mathematics
1 answer:
Viefleur [7K]3 years ago
7 0

Answer:

how old are the cats that live in Kyle's neighborhood?

Step-by-step explanation:

Because the question is  statistical which mean you will get more then one answers

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Which graph corresponds to the following inequality? A. B. C. D.
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I can see the question but if you add a picture maybe i’ll understand it a little more
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An organization will give a prize to a local artist. The artist will be randomly chosen from among 7 painters, 4 sculptors, and
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The probability would be 12/16 or 75%.
Hope this helps!
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3 years ago
Simplify 4 + (-3) - 2 x (-6 )
sukhopar [10]
If you would like to simplify <span>4 + (-3) - 2 * (-6), you can do this using the following steps:

</span><span>4 + (-3) - 2 * (-6) = 4 - 3 + 2 * 6 = 1 + 12 = 13
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The correct result would be 13.
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3 years ago
HELP DUE IN 15 MINS!
Mamont248 [21]

Answer:

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Step-by-step explanation:

8 0
3 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, -90^{\circ}

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

  • \cos \theta = \cos(-\theta)
  • \sin(-\theta) = -\sin \theta

\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

Therefore, the degree of rotation is, -90^{\circ}

7 0
3 years ago
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