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This is a linear graph. When the y value is -0.007, I cannot get the value of x that should be on the trendline and propostional to the y value. It is far apart from the line. Is it something wrong with setting of the excel? Pls help me!!!!!
This is a linear graph. When the y value is -0.007, I cannot get the value of x that should be on the trendline and propostional to the y value. It is far apart from the line. Is it something wrong with setting of the excel? Pls help me!!!!!
Last edited by JennyLaw; 02-04-2009 at 12:54 AM. Reason: Re- attach file
Welcome to the forum.
There isn't a formula or graph in your workbook. Maybe you'll give us a clue as to what you're talking about.
Entia non sunt multiplicanda sine necessitate
Graph is as attachment.
There is no graph in the attachment.
Sorry, i attach the wrong file. very sorry
I replace the attachment with the right file already.
OK, I see the graph. So what's your question?
In B7, =-0.007
In A7, =TREND(A2:A6, B2:B6, B7) ~ 331.14
the question is that when i replace the y value with - o.oo7, the x value that i get is 337.44 and it is not on the trendline. In my opinion, the x value supposed at around 331 and should be less than the highest value plotted in the graph.
How can u get that? could u please explain to me?
why is the trend line given by the excel doesn't give me a correct value?
if i place
y = 0.0025x - 0.8506
when Y = -0.007, the formular given by the trend line will give me a value of 337.44 instead of the correct value of 331.14
is that some setting i must specifiy in formating the trend line?
Last edited by JennyLaw; 02-04-2009 at 01:41 AM.
That's a rounding issue.
Increase the decimal placage on the trend line data label.
Re: Validation of equation on chart
Thank you very much. The problem was solved.
Last edited by JennyLaw; 02-04-2009 at 09:34 PM.
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