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igomit [66]
3 years ago
6

Causal models a. provide evidence of a causal relationship between an independent variable and the variable to be forecast. b. u

se the average of the most recent data values in the time series as the forecast for the next period. c. relate a time series to other variables that are believed to explain or cause its behavior. d. occur whenever all the independent variables are previous values of the same time series.
Mathematics
1 answer:
marshall27 [118]3 years ago
7 0

Answer:A. provide evidence of a causal relationship between an independent variable and the variable to be forecast

Step-by-step explanation: Casual model tends to show the cause and effect relationship between the dependent variable to be forcasted and the independent variables upon which the dependent variable is dependent.

Casual model is frequently used in the field of Statistics and Economics when making forcasts about future investments or the cause of certain events,knowing what activities to carry out in the future.

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Find the solution to the linear system by using the x - and y -intercepts.
Elenna [48]

The solution to these equations is (x, y) = (2, 4).

We've not heard of a method of solving a linear system using the x- and y-intercepts. A Google search on the subject turns up this question, and no other information. However, there is information you can gain from the intercepts that can help you find a solution. (Your reference material may provide a better source of information on this subject.)

The <em>intercept form</em> of a linear equation is ...

... x/(x-intercept) + y/(y-intercept) = 1

Dividing the first equation by 4, you can rearrange it to ...

... x/(-2) +y/(2) = 1 . . . . . . the x-intercept is -2, the y-intercept is +2.

Dividing the second equation by -6, you can rearrange it to ...

... x/-6 +y/3 = 1 . . . . . . . . the x-intercept is -6, the y-intercept is +3.

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<em>What you can do with the intercepts</em>

The intercepts can be used to <em>graph the equations</em>. Plot each of the intercepts for a given equation, then draw a line through them. (See the attachment.)

Here, both lines have their intercepts on the negative x and positive y axes. The slopes and intercepts of the lines are such that they intersect in the 1st quadrant.

We can use the intercepts to <em>find the slope of the line</em>.

The slope of each line can be found from ...

... slope = -(y-intercept)/(x-intercept)

Then the slope of the first line is ...

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and the slope of the second line is ...

... m2 = -3/-6 = 1/2

The difference in slopes is ...

... m1 - m2 = 1 - 1/2 = 1/2

Using the slope and intercept we can <em>find the solution</em> by substitution or elimination.

For line 1 with slope m1 and y-intercept b1, the equation of the line in slope-intercept form is

... y = m1·x + b1 = x +2

For line 2 with slope m2 and y-intercept b2, the equation of the line in slope-intercept form is

... y = m2·x +b2 = (1/2)x +3

Subtracting the second equation from the first eliminates the y-variable and gives ...

... y - y = 0 = (x +2) -(1/2x +3) = 1/2x - 1

Adding 1 and multiplying by 2 gives the solution for x:

... x = 2

Then the first equation gives the solution for y:

... y = x + 2 = 2+2 = 4

The solution is (x, y) = (2, 4).

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<em>Comment on </em><em>find the solution</em>

Above, we subtracted one equation from the other to get ...

... y - y = 0 = x(m1 -m2) +(b1 -b2)

We could have simply equated the values of y to get ...

... m1·x +b1 = y = m2·x +b2

Either way you do this, you find the solution for x is ...

... x = (b2 -b1)/(m1 -m2)

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