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A graph shows the survey results for a group of students who were asked how many honors classes they have taken and how many elective classes: A scatter plot is shown with the title class choices. The x axis is labeled number of honors classes and the y axis is labeled number of electives. Data points are located at 1 and 8, 3 and 6, 3 and 9, 5 and 3, 6 and 6, 6 and 9, 8 and 6. A line of best fit crosses the y axis at 9 and passes through the point 6 and 6. How many elective classes would students likely have taken if they have taken 12 honors classes? 15, because y = one halfx + 9 12, because y = y = negative one halfx + 9 6, y = ˜one halfx + 9 3, because y = negative one halfx + 9

Respuesta :

carlosego
The first thing to do in this case is to find the equation of the line that represents the problem.
 You have two points given:
 (0, 9)
 (6, 6)
 The generic equation is:
 y-yo = m (x-xo)
 Where;
 m = (y2-y1) / (x2-x1)
 m = (6-9) / (6-0)
 m = -3 / 6
 m = -1 / 2
 We choose any point:
 (xo, yo) = (0, 9)
 We rewrite:
 y-yo = m (x-xo)
 y-9 = (- 1/2) (x-0)
 y = -1 / 2x + 9
 So we have that for x = 12
 y = -1 / 2 (12) +9
 y = -6 + 9
 y = 3
 answer:
 3, because y = negative one halfx + 9

Answer:

Option D.

Step-by-step explanation:

It is given that a line of best fit crosses the y axis at 9 and passes through the point (6,6).

If a line passes through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the equation of line is

[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

The line passes through (0,9) and (6,6). So, the equation of line is

[tex]y-9=\frac{6-9}{6-0}(x-0)[/tex]

[tex]y-9=-\dfrac{1}{2}(x)[/tex]

Add 9 from both sides.

[tex]y=-\dfrac{1}{2}(x)+9[/tex]

The equation of best fit line is [tex]y=-\dfrac{1}{2}(x)+9[/tex].

Substitute x=12 in the above equation.

[tex]y=-\dfrac{1}{2}(12)+9[/tex]

[tex]y=-6+9[/tex]

[tex]y=3[/tex]

3 elective classes would students likely have taken if they have taken 12 honors classes.

Therefore, the correct option is D.

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