Question

The equation for these two graphs below.
(I just need help putting it all together, I have the X and Y points just everything else is confusing me)

Answer 1

I chose the points 33 46.23 and 1050 49.74...
\[\frac{ 49.74 - 46.23 }{ 1050 - 33 }\]

If i am correct thats the correct way?

Answer 2

This \[\frac{ 49.74-46.23 }{ 1050-33 }\] would come out to be \[\frac{ 3.51 }{ 1017 }\] if i am correct.

Answer 3

3.51 divided by 1017 should come out to be 0.0034513274336283 and that would be m correct?
Cause the equation is y=mx+b

Answer 4

I'm making my own equation based off of the example below. I got this far and im now stuck..

Answer 5

One thing I notice is that you haven't provided the instructions that apply to this problem. Would you mind doing that?

Answer 6

Sure thing just give me a moment or two

Answer 7

"putting it all together" just isn't specific enough.

Answer 8

From the graphs i have to make a linear equation from start to finish. (The graphs are the same just different style). I have gotten to the point above and I'm now stuck on what else to do. I have the points picked out as seen, i figured everything out. Basically I'm making the equation from head to toe.


Is this good enough. Its kind of hard to explain.

Answer 9

1) You need to draw a straight line thru the given point plots "by eye." The left end of your line appears to be placed more or less correctly. The right end of your line appears to be too high; move it down a bit.

2) Extend this straight line to cross both the x- and the y-axes. Identify the coordinates of the intercepts. does this language sound familiar to you?

Answer 10

Not really... and the way the graphs are made there isnt a possible way to extend them..

Answer 11

y=mx+b
I think i have found the y=mx part now i just need to find b if i am correct...I cant remember..Sorry for taking so much of your time..

Answer 12

Then you'll have to approximate points on the line and determine the slope and y-intercept of the line from them.

As before, I'd suggest that you move the right end of your line down a bit.

But to discuss procedure, let's just use the line you already have.

supposing that x=0, y is approx. 39. Do you agree or disagree?

Answer 13

Looking at both graphs i agree.

Answer 14

In other words: Draw an imaginary vertical line thru 7000 on the x-axis. Find the point at which this line intersects your graph. Using a ruler or the straight edge of a piece of paper, determine the approx. y-coordinate of this point of intersection.

Answer 15

the points would possibly be (14.16, 40)? I'd say the approx Y coordinate is either 39 or 40....Even with drawing a line thru it it is confusing me. (im doing all this on my laptop)

Answer 16

@mathmale (sorry. forgot to mention you in the previous message)

Answer 17

One point would be (0,39). The other would be (7000, what y value?) Again, determine the y-component of the point at which your line and the vertical line thru x=7000 intercept.

Answer 18

(7000, 40) looks to be it when i place it... @mathmale

Answer 19

sorry not 40. (7000, 44) maybe... i just redid it...

Answer 20

Please compare your result (7000,44) to mine: (7000, 26). Note that these are all approximations.

Why 44 as your y-component?

Answer 21

When i did it i put mine up near the line that they made. I positioned it around 44 so thats why i thought 44. I graphed yours and that seems much better..

Answer 22

Supposing you're willing to accept that the coordinates of 2 points on your line are (0,39) and (7000,26)"

The slope of your line would be \[m=\frac{ 26-39 }{ 7000-0 }=\frac{ -13 }{ 7000 }\]

Answer 23

ok now we would divide -13 by 7000 correct? if so... it'd be -0.0018571428571429? Now if we needed to round that it'd be -0.00186? @mathmale

Answer 24

I have to write an equation out for it in the form of the y=mx+b.....

Answer 25

You now have a value for the (negative) slope of your regression line. You have 2 points on this line. Choose one of them (it doesn't matter which one).

the "point slope form for the equation of a straight line" is \[y-y _{0}=m(x-x _{0})\]

Answer 26

substitute your value for m into this equation. Also substitute the coordinates of the point you chose into the equation, replacing \[(x _{0},y _{0}).\]

Answer 27

Simplify. Re-write your equation in the form y=mx+b.

Answer 28

I dont know how to put it into the "point slope form for the equation of a straight line" form....

Answer 29

26 = (-0.00186) x (0) = 0
26=0+b
26 + 0 = 0+0+b
26=b
y= -0.00186x + 26???

@mathmale

Answer 30

Let's look at \[y-y _{0}=m(x-x _{0}).\]

Answer 31

Substitute -13/7000 for m. Assuming that you've chosen to work with the point (7000,26), subs. 7000 for x_0 and 26 for y_0.

Answer 32

that'd come out differently from your result:\[y-26=\frac{ -13 }{ 7000 }(x-7000)\]

Answer 33

Could you solve that for y? You might want to write this equation as \[y-26=\frac{ -13 }{ 7000 }(x-7000)=\frac{ -13 }{ 7000 }x + 13~first.\]

Answer 34

Add 26 to both sides of the equation.

Answer 35

for the right side 12.95171428?

Answer 36

if i do the left
-13 divided 7000 (26-7000) it comes out the same.... I'm confused on what to do @mathmale