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evablogger [386]
3 years ago
11

What is the overlap of data set 1 and data set 2 A. High B.moderate C.low D.none

Mathematics
1 answer:
finlep [7]3 years ago
7 0
It's definitely a low data overlap. It's not medium because there's not a solid amount of matching data, and its not 'none' because there is matching data.
so it’s C.
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You would factor a 5x out of the numerator sine there is a 5x in 5x^3, 10x^2 and 15x. when you factor, it should be 5x( x^2+2x+3). The 5x on top and bottom would then cancel out, leaving x^2+2x+3 as you answer.
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I assume you know about the dot product, and that for two vectors \mathbf a and \mathbf b, the angle between them \theta satisfies

\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}

Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make \cos\theta=1 or \cos\theta=-1, respectively.

Part A)

\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2

\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2

\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4

Then the angle between \vec v_1,\vec v_2 is such that

\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}

so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).

Part B) involves the same computations:

\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}

\vec u_2 has the same components but differing by sign and order, as \vec u_1; its magnitude remains the same, though:

\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}

\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12

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