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3 years ago

2y+-1y+-1y+3=-13 solve for y

Mathematics

2 answers:

olasank [31]3 years ago

7 0

Answer:

these problem can't be solve because no y remains here because of subtraction of 2y-2y

jeka57 [31]3 years ago

4 0

Answer:

There are no values of y

that make the equation true.

No solution

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3 years ago

24 H in a d? 90 D in a R A?

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Answer:

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Step-by-step explanation:

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4 years ago

Find the missing length. the triangles are similar.

stich3 [128]

70 + 78 + 91 = 239
239 - 42 - 36 = 161

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3 years ago

Read 2 more answers

In a two-player game, five cards, numbered 1 through 5, are placed in a bag. A card is drawn at random, and the players look at

IceJOKER [234]

Answer:

If it is less than 3, Player 1 earns 3 points.

If not, Player 2 earns 2 points.

Step-by-step explanation:

Player 1 :

p(N < 3) = p(N = 1 or N = 2) = 2/5

Player 2 :

p(N ≥ 3) = p(N = 3 or N = 4 or N = 5) = 3/5

We notice that :

p(N < 3) × 3 = (2/5) × 3 = 6/5

On the other hand,

p(N ≥ 3) × 2 = (3/5) × 2 = 6/5

since ,the probability player 1 win multiplied by the associated number of points (3)

is equal to

the probability player 2 win multiplied by the associated number of points (2).

Then the game is fair.

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2 years ago

Read 2 more answers

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