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4 years ago

PLS HELP graph y=−x+6 8x−4y=−16

Mathematics

1 answer:

Vlad1618 [11]4 years ago

7 0

Answer:

The graph is shown below.

Step-by-step explanation:

Given:

$y=-x+6\\8x-4y=-16$

In order to graph these two lines, we find their x and y intercepts.

The x intercept of $y=-x+6$ is at $y=0$ . So,

$0=-x+6\\x=6$

Therefore, x intercept is at $(6,0)$

The y intercept of $y=-x+6$ is at $x=0$ . So,

$y=0+6\\y=6$

Therefore, y intercept is at $(0,6)$

Now, similarly, we find the x and y intercepts of the other line.

The x intercept of $8x-4y=-16$ is at $y=0$ . So,

$8x-4(0)=-16\\8x=-16\\x=\frac{-16}{8}=x=-2$

Therefore, x intercept is at $(-2,0)$

The y intercept of $8x-4y=-16$ is at $x=0$ . So,

$8(0)-4y=-16\\-4y=-16\\y=frac{-16}{-4}=y=4$

Therefore, y intercept is at $(0,4)$

Now, we plot these pair of points on a graph and then join the pairs of lines.

The graph is shown below.

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The difference between the observed value of the dependent variable and the value predicted using the estimated regression equat

Elenna [48]

Answer:

For this case we define the dependent variable as Y and the independent variable X. We assume that we have n observations and that means the following pairs:

$(x_1, y_1) ,....,(x_n,y_n)$

For this case we assume that we want to find a linear regression model given by:

$\hat y = \hat m x +\hat b$

Where:

$\hat m$ represent the estimated slope for the model

$\hat b$ represent the estimated intercept for the model

And for any estimation of the dependent variable $\hat y_i , i=1,...,n$ is given by this model.

The difference between the observed value of the dependnet variable and the value predicted using the estimated regression equation is known as residual, and the residual is given by this formula:

$e_i = y_i -\hat y_i , i=1,...,n$

So the best option for this case is:

d. residual