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timurjin [86]
3 years ago
11

If events X and Y are independent, what must be true? Check all that apply. P(Y | X) = 0 P(X | Y) = 0 P(Y | X) = P(Y) P(Y | X) =

P(X) P(X | Y) = P(Y) P(X | Y) = P(X)
Mathematics
2 answers:
lana [24]3 years ago
3 0

if events X and Y are independent, then for intersection we multiply the probability

P(Y∩X)  = P(Y) * P(X)

We know that

P(Y|X) =\frac{P(YintersectionX)}{P(X)}

Now we replace P(Y) * P(X) for  P(Y∩X)

P(Y|X) =\frac{P(Y)*P(X)}{P(X)}

Cancel out P(X)

So P(Y|X) = P(Y)

Like that

P(X|Y) =\frac{P(XintersectionY)}{P(Y)}

Now we replace P(X) * P(Y) for  P(X∩Y)

P(X|Y) =\frac{P(X)*P(Y)}{P(Y)}

Cancel out P(Y)

So P(X|Y) = P(X)

P(Y | X) = P(Y)  and P(X | Y) = P(X)  are true



melamori03 [73]3 years ago
3 0

Answer:

3 and 6 are correct

Step-by-step explanation:

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Answer: 5/6

Step-by-step explanation:

2/3=4/6

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5 0
3 years ago
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(5a^2+6a+2)-(7a^2-7a-5)
krok68 [10]

For this case we must simplify the following expression:(5a^ 2 + 6a + 2) - (7a^ 2-7a-5) =

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- * + = -\\- * - = +

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5a ^ 2 + 6a + 2-7a ^ 2 + 7a + 5 =

We add similar terms taking into account that:

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8 0
4 years ago
A can of tuna has a volume 18pi cm3 and a height of 2 cm. Find the area of the label that wraps around the entire can and does n
Sphinxa [80]

Answer:

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Step-by-step explanation:

step 1

Find the radius

we know that

The volume of the cylinder (can of tuna) is equal to

V=\pi r^{2} h

In this problem we have

V=18\pi\ cm^{3}

h=2\ cm

substitute and solve for r

18\pi=\pi r^{2} (2)

Simplify

18=r^{2}(2)

r^{2}=9

r=3\ cm

step 2

<em>Find the lateral area of the can</em>

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LA=2\pi rh

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r=3\ cm

h=2\ cm

LA=2\pi (3)(2)=12\pi\ cm^{2}

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igor_vitrenko [27]
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now divide both terms in V by that:

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see attached picture: 

7 0
3 years ago
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