Would I be right in assuming that on a scatter graph with a linear trendline, the equation (using display equation on the linear trendline) would be able to be used to determine what X would be at a known Y?
My equation is this Y = 0.0091x - 0.3865
So be re-arranging it I would get X=Y/0.0091+0.3865, does this look right?
Graph Help please!?
2017-02-05 4:14 am
回答 (4)
2017-02-05 4:23 am
✔ 最佳答案
So be re-arranging it I would get X=Y/0.0091+0.3865, does this look right?Not to me !!
y = 0.0091x - 0.3865
solving for x in terms of y...
x = (y + 0.3865) / 0.0091
You screwed-up order of operations !!
2017-02-05 4:24 am
NO, that would not be right.
Y = 0.0091X − 0.3865
0.0091X = Y + 0.3865
X = (Y + 0.3865)/0.0091
X = Y/0.0091 + 0.3865/0.0091
You must divide ALL of (Y + 0.3865) by 0.0091, not just Y
2017-02-05 4:49 am
Thanks guys haha, honestly, as you can probably tell I have no idea what I am doing here. so using the formula that you have stated I will get the correct X value for a known Y?
Say Y=10 then X=1141.373626?
or if Y=20 then X=2240.274725?
and so on
Say Y=10 then X=1141.373626?
or if Y=20 then X=2240.274725?
and so on
2017-02-05 4:37 am
Not quite right. The 0.3865 also needs to be divided by 0.0091:
.. X = (Y +0.3865)/0.0091
.. X = (Y +0.3865)/0.0091
收錄日期: 2021-04-24 00:12:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170204201422AASW4sM