The scatter plot below shows the profit earned each month by a new company over the first year of operation. the owner writes a line of best fit equation, shown below, to model the relationship between profit earned and month. y = 2,500x - 2,500 explain how you know that the line of best fit equation is appropriate, mentioning both the slope and y-intercept in your response.

The regression line fits the data well. Problem can be when there is a trend, like exponential growth, but someone still wants to use linear regression. Although in this case it seems to be okay. In other words, the data points can be found under or over the line randomly. In a problematic case is when for example, there are 5 data points next to each other under the line. Usually companies don't make profit when they're first starting.

First  I focused on 3 or more points that were collinear to each other.  Collinear means that when you connect these points, you get a straight line.  These points are at months 1, 2, 3, and 9.  Then of these points, I chose 2 points to find the slope between them.

Point 1 = (1, 0)

Point 2 = (2, 2.5) or  (2, 2500)         because profit is measured in thousands

These are the first two points.

Slope = (2500 - 0) / (2 - 1)

Slope = 2500 / 1

Slope = 2500

Then I plug in the slope and values of point 1 into the slope-intercept form of the line.

y = mx + b

0 = 2500(1) + b

Now we have to Solve for b.

0 = 2500 + b

-2500 = b

Plug in this value of b into the slope intercept form of the equation to get the best fit line.

 y = 2500x - 2500

x = -2

y = 848

step-by-step explanation:

plug in the numbers

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5