Question

Can someone help me find the equation to this graph? @radar @whpalmer4

Answer 1

I started to draw on the graph but i didnt know if that was correct.... i was looking for an equation

Answer 2

Gee, that looks like a lab question. You need to fit a line on that drawing that fits that pattern, you're line will be a straight one. Then using the values on that graph calculate the slope. and intercept.

Answer 3

The equation is for you to develop from your drawn line. It is like a practical exercise.

Answer 4

thats what i did.

Answer 5

Lay a straight edge, aligned with the concentration of scatter points, then select two points and doe the change in y over the change in x thing giving you the slope. Do you have to turn in the drawing?

Answer 6

yeah, I have to draw the equation then come up with an equation... which is what I'm struggling with- i dont know how to come up with the eqaution

Answer 7

Good, I would select two points, I chose (22,28) and (0,6) then m or slope would be:
(28-6)/(22-0) or 22/22 or 1, The y intercept when (x=0) looks like 6
My equation would then be y=x + 6

Answer 8

Now you can do something similar, selecting two points that makes it easy for you.

Answer 9

I did use your line however.

Answer 10

its kind of hard to do that when they're decimal points

Answer 11

Not really, look at the points I selected if you follow the grid line where y = 28 over to the point where it intersects your line, now drop straight down and you will hit the point 22 on the x axis, or at least I did. lol

Answer 12

I then chose the point (0,6), easy cause 0 is easy to find and go up to where your line is it is midway between 4 and 8 on the y axis, I chose (0,6)

Answer 13

No decimals.

Answer 14

As far as drawing the line goes, I usually try to find a line that has approximately the same number of points above it as below it. Your line has a handful that are just a smidge below, tipping the balance slightly, but I think you captured the trend quite well. And you just eyeball the x,y values at the ends of your trend line (better than picking points in the middle of the line, unless they go exactly through grid intersections), no need to kill yourself — if you were really needing a very good fit for the line, you'd use some software that captured all the data points for you and made the best-fit line. One technique for doing that is called "least squares", where you come up with a line that minimizes the sum of the squares of the distance from the line to each data point, effectively finding you a line that goes right through the middle of the data. It's not something you would want to work out by hand!

Answer 15

Through the magic of the Mk I eyeball, I scanned your data from the scatter plot and made my own copy, then had software calculate a best-fit approximation. The equation for the line it found is \[y = 0.868732x+6.89742\]which is pretty close to what what my pal @radar did with much less effort :-)

Here's a graph showing the data and that fitted line:

Answer 16

I tend to have trouble getting OpenStudy to show me pages around this time of the day, so I wasn't able to respond earlier. Hopefully the party gifts I brought made up for the late arrival :-)

Answer 17

Glad I got that close using Kentucky windage. ):

Answer 18

I admit I didn't draw the line, but just used it. The poster drew the line, it looked good enough to use.

Answer 19

Quite a bit of Kentucky windage in my eyeballing of the data points! A real Mathematica geek would have written some code that analyzed the graphic directly instead of my estimating x,y for each of those 24 points on a small graph, but I'm not a real Mathematica geek, and don't even play one on TV :-)

Answer 20

Fine job, you have a nice day.