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)?(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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