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hjlf
4 years ago
5

3.2 divided by 8 what do I get?

Mathematics
1 answer:
zepelin [54]4 years ago
8 0
The answer will be 0.4
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6 0
3 years ago
Please help I will reward brainly!!! Need done ASAP
Tomtit [17]
To solve the question we use the following analogy:
Given that the odds of winning is a/b, then the probability is b/(a+b)
hence to solve the question we proceed as follows:

a]
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odds against E=(36-11)/11
simplifying this gives us:
25/11

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simplifying this we get:
5/31
hence odds in favor of E=31/5


c]P(E)=32/36
odds in favor of E will be:
36/(36-32)
simplifying the above we get:
36/4
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(36-30)/36
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6 0
4 years ago
Assume ​Y=1​+X+u​, where X​, Y​, and ​u=v+X are random​ variables, v is independent of X​; ​E(v​)=0, ​Var(v​)=1​, ​E(X​)=1, and
kobusy [5.1K]

Answer:

a) E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+

b) E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2

c) E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2

d) E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6

e) E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1

f) E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3

g) E(u) = E(v) +E(X) = 0+1=1

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

Y = 1+X +u

u = v+X

with both Y and u random variables, we also know that:

[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2

And we want to calculate this:

Part a

E(u|X=1)= E(v+X|X=1)

Using properties for the conditional expected value we have this:

E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1

Because we assume that v and X are independent

Part b

E(Y| X=1) = E(1+X+u|X=1)

If we distribute the expected value we got:

E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2

Part c

E(u|X=2)= E(v+X|X=2)

Using properties for the conditional expected value we have this:

E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2

Because we assume that v and X are independent

Part d

E(Y| X=2) = E(1+X+u|X=2)

If we distribute the expected value we got:

E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6

Part e

E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1

Part f

E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3

Part g

E(u) = E(v) +E(X) = 0+1=1

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

8 0
3 years ago
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