[R] Unsigned Posts; Was Setting graphical parameters

[Bert Gunter]

Folks:

Rolf's (appropriate, in my view) response below seems symptomatic of an
increasing tendency of posters to hide their identities with pseudonyms and
fake headers. While some of this may be due to identity paranoia (which I
think is overblown for this list), I suspect that a good chunk of it is lazy
students trying to beat the system by having us do their homework. The
etiquette of this list has traditionally been, like the software, open: we
sign our names. I would urge helpeRs to adhere to this etiquette and ignore
unsigned posts.

Contrary views welcome, either via public or private responses.

-- Bert

Bert Gunter
Genentech Nonclinical Statistics

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of Rolf Turner
Sent: Wednesday, March 03, 2010 6:42 PM
To: Pitmaster
Cc: r-help at r-project.org
Subject: Re: [R] Setting graphical parameters


Do your own <expletive deleted> homework!

It looks pretty trivial; what's your problem?

	cheers,

		Rolf Turner

On 4/03/2010, at 2:39 PM, Pitmaster wrote:

> 
> Hi guys... I have problem with this excersise... 
> 
> Consider the pressure data frame again.
> 
> (a) Plot pressure against temperature, and use the following
> command to pass a curve through these data:
> 
>> curve((0.168 + 0.007*x)ˆ(20/3), from=0, to=400, add=TRUE)
> 
> (b) Now, apply the power transformation y3/20 to the pressure data values.
> Plot these transformed values against temperature. Is a linear
> or nonlinear relationship evident now? Use the abline() function
> to pass a straight line through the points. (You need an intercept
> and slope for this – see the previous part of this question to obtain
> appropriate values.)
> 
> (c) Add a suitable title to the graph.
> 
> (d) Re-do the above plots, but use the mfrow() function to display
> them in a 2 × 1 layout on the graphics page. Repeat once again
> using a 1 × 2 layout.
> 
> DATA:
>> pressure
>   temperature pressure
> 1            0   0.0002
> 2           20   0.0012
> 3           40   0.0060
> 4           60   0.0300
> 5           80   0.0900
> 6          100   0.2700
> 7          120   0.7500
> 8          140   1.8500
> 9          160   4.2000
> 10         180   8.8000
> 11         200  17.3000
> 12         220  32.1000
> 13         240  57.0000
> 14         260  96.0000
> 15         280 157.0000
> 16         300 247.0000
> 17         320 376.0000
> 18         340 558.0000
> 19         360 806.0000
> 
> 
> Can anyone know the solution ?

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