On Sep 4, 2014, at 8:40 AM, Basilius Sapientia wrote:
> I have this code:
> Vm <- c(6.2208, 4.9736, 4.1423, 3.1031, 2.4795, 1.6483, 1.2328, 0.98357,
> 0.81746, 0.60998); #Molvolume
> p <- c(0.4, 0.5, 0.6, 0.8, 1, 1.5, 2, 2.5, 3, 4)*1000; #Pressure
> Rydb <- 8.3144621; #Constant
> Tempi <- 300; #T
You will find lots of examples if you do an internet search for
R quadratic regression
Here's just one ... http://www.theanalysisfactor.com/r-tutorial-4/
Jean
On Thu, Sep 4, 2014 at 10:40 AM, Basilius Sapientia
wrote:
> I have this code:
> Vm <- c(6.2208, 4.9736, 4.1423, 3.1031, 2.4795, 1
I have this code:
Vm <- c(6.2208, 4.9736, 4.1423, 3.1031, 2.4795, 1.6483, 1.2328, 0.98357,
0.81746, 0.60998); #Molvolume
p <- c(0.4, 0.5, 0.6, 0.8, 1, 1.5, 2, 2.5, 3, 4)*1000; #Pressure
Rydb <- 8.3144621; #Constant
Tempi <- 300; #Temperature in Kelvin
Vmi <- Vm^(-1); #To get Vm^(-1)
Zi <- (p*Vm)/(
.
~ John Tukey
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens JoeP
> Verzonden: dinsdag 26 juli 2011 11:26
> Aan: r-help@r-project.org
> Onderwerp: [R] help with regression
>
> Hi,
>
> I am trying to
Hi,
I am trying to do a linear regression but I want one of my variables to not
generate a coefficient. E.g. what I want to do is fit for y=a+b+c but
forcing the coefficient of b to be 1. Is this possible?
I have been fitting y-b=a+c but I have found that when I recalculate y it is
not close
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