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3 years ago

Y=x+2 y+1= – 4x Is this ordered pair a solution of the system of equations? Help:(

Mathematics

1 answer:

Shkiper50 [21]3 years ago

3 0

y=x+2.....(1)
y+1= – 4x......(2)

Put value of y in Equation 1 in Equation 2.

$\looparrowright \sf y + 1 = - 4x$

$\\ \\$

$\looparrowright \sf (x + 2) + 1 = - 4x$

$\\ \\$

$\looparrowright \sf x + 2 + 1 = - 4x$

$\\ \\$

$\looparrowright \sf x +3= - 4x$

$\\ \\$

$\looparrowright \sf x + 4x= - 3$

$\\ \\$

$\looparrowright \sf 5x= - 3$

$\\ \\$

$\looparrowright \sf x= \dfrac{ - 3}{5}$

$\\ \\$

Let's find value of y :

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$\looparrowright \sf y=x+2$

$\\ \\$

$\looparrowright \sf y= \dfrac{ - 3}{5} +2$

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$\looparrowright \sf y= \dfrac{ - 3 + 10}{5}$

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$\looparrowright \sf y= \dfrac{7}{5}$

Solution:

$\looparrowright \sf \bigg( \dfrac{ - 3}{5} , \dfrac{7}{5} \bigg)$

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Step-by-step explanation:

If whats-his-face can do it in 4 and girl can do it in 2 divide 4 and 2 and get 2.

Hope you get it right tell me if this is correct or not.

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3 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

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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]

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Hence, 4 ounces of oil will be used for the soup.

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