3 years ago

Graph 16x−4y=2.Please answer I will make you the brainlist and rate you 5 stars if good.And report if bad.

Mathematics

2 answers:

KonstantinChe [14]3 years ago

3 0

Answer:

First one is the graph second one is the equation

Step-by-step explanation:

Hope this helps :)

solong [7]3 years ago

3 0

Answer:

Step-by-step explanation:

as graphed

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Step-by-step explanation:

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3 years ago

G with the definition of covariance, prove cov[(ay − b),(cy − d)] = accov(x, y ), where x, y are random variables and a, b, c, d

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By definition of covariance,

$\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]$

$\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]$

We have

$\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]$

$=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd$

$\mathbb E[aX-b]=a\mathbb E[X]-b$

$\mathbb E[cY-d]=c\mathbb E[Y]-d$

$\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd$

Putting everything together, we find the covariance reduces to

$\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)$

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3 years ago

Write twelve and five hundred ninety nine thousandths in standard form

stellarik [79]

12.599
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8 pounds

Step-by-step explanation:

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