Find four solutions of the equation. Write the solutions as ordered pairs. 3x-y=-6

Mathematics

1 answer:

Nezavi [6.7K]1 year ago

3 0

<h2>Key Ideas</h2>

Linear equations
Finding solutions to the linear equation

<h2>Solving the Question</h2>

We're given:

3x-y=-6
Find 4 solutions

We can start by isolating one of the variables, like <em>y</em>:

$3x-y=-6\\-y=-3x-6\\y=3x+6$

Now we can find four solutions by plugging in any number for <em>x</em>:

$y=3x+6\\y=3(1)+6\\y=3+6\\y=9$

First solution: (1,9)

$y=3x+6\\y=3(2)+6\\y=6+6\\y=12$

Second solution: (2,12)

$y=3x+6\\y=3(3)+6\\y=9+6\\y=15$

Third solution: (3,15)

$y=3x+6\\y=3(4)+6\\y=12+6\\y=18$

Fourth solution: (4,18)

<h2>Possible Answers</h2>

First solution: (1,9)

Second solution: (2,12)

Third solution: (3,15)

Fourth solution: (4,18)

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If X and Y are any random variables with E(X) = 5, E(Y) = 6, E(XY) = 21, V(X) = 9 and V(Y) = 10, then the relationship between X

natka813 [3]

Answer:

We have a strong negative relationship between the variables.

Step-by-step explanation:

Given two random variables X and Y, it is possible to calculate the covariance as Cov(X, Y) = E(XY)-E(X)E(Y). We have E(X)=5, E(Y)=6 and E(XY)=21. Therefore Cov(X,Y)=21-(5)(6)=21-30=-9. On the other hand, we know that the correlation of X and Y is the number defined by $Cov(X,Y)/\sqrt{Var(X)}\sqrt{Var(Y)}$ and because in this particular case we have V(X)=9 and V(Y)=10, we have $-9/\sqrt{9}\sqrt{10}$ = -0.9487. Therefore, we have a strong negative relationship between the variables.