I\'m trying to solve problem 11.7 but don\'t understand how to solve with more t

Question

I'm trying to solve problem 11.7 but don't understand how to solve with more than one variable.

Similarly, if the line passes through the point (4.6), th x = 4, y = 6 must satisfy the equation: that is, 11.4 If a straight-line probabilistic relationship relates the mean E(v) to an independent variable x, does it imply that every value of the variable y will always fall exactly on the line of means? Why or why not? 6 = 0 + 1(4) Learning the Mechanics 11.5 In each case, graph the line that passes through the givern Use these two equations to solve for and !; then find the equation of the line that passes through the points (-2, 4) and (4,6). points a. (1, 1) and (5,5) b. (0,3) and (3, 0) c. (1, 1) and (4, 2) d. (-6,-3) and (2,6) 11.8 Refer to Exercise 11.7. Find the equations of the linesthag pass through the points, listed in Exercise 11.. 11.9 Plot the following lines: lins rsptedf a.), = 4 + x b, y = 5-2x 11.7 The equation (deterministic) for a straight line is 2x r. y=.50 + 1.5x If the line passes through the point (-2,4), then -2, y-4 must satisfy the equation; that is, 1.10 CGive the slope and y-intercept for cach of he nisadiad in Exercise 11.9

Explanation / Answer

We have 2 variable and two equations:

1) 4 = bo + b1 * - 2

2) 6 = bo + b1 * 4

Subtract these equations:

2 = 6 * b1

b1 = 2/6 = 0.33

Now put this in 6 = bo + b1 * 4

6 = bo + 0.33 * 4

bo = 4.67

So, equation is:

y = 4.67 + 0.333 x

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