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Hi all, I am wondering what people are doing for security when registering IP phone's remotely if you do not have the equipment to do a VPN tunnel at the remote site. The phone I would be working with mainly is the Polycom lineup. Thanks, Connor Spiess -------------- next part -------------- An HTML attachment was scrubbed... URL:

I am using R on a Windows XP professional platform. The following code is part of a bigger one CODE press=function(y,x){ library(qpcR) models.press=numeric(0) cat("\n") dep=y print(dep) indep=log(x) print(indep) yfit=dep-PRESS(lm(dep~indep))[[2]] cat("\n yfit\n") print(yfit) yfit.orig=yfit presid=y-yfit.orig press=sum(presid^2)

Dear R-helpers, does anyone know of some package/function that can build a network from the functions that are implemented in a package, i.e. visualize the cross-references from one function to another in the same or some dependent package? An example would be a function like 'nls' on top of the hierarchy and then a network of nodes from the functions that are called within 'nls'

Dear R-helpers, I´m in the context of writing a general function for error propagation in R. There are somehow a few questions I would like to ask (discuss), as my statistical knowledge is somewhat restricted. Below is the function I wrote, the questions are marked. Many thanks in advance. propagate <- function(expr, varList, type = c("stat", "raw"), cov = TRUE) {

Ich werde au?er Haus sein von 18-06-99 Bis 28-06-99. Ich werde Ihre Nachrichten nach meiner R?ckkehr beantworten. Bitte wenden Sie sich in dringenden F?llen an: Frank R?seler, Tel: MON 3855 Uwe Spiess, Tel: MON 3743 Guido Zimmermann, Tel: MON 3962 Ferdinand Gro?, Tel: MON 2368 lieber Gru? Axel Hellmig

...sten servieren. Mit Erdbeeren aus der K?hltruhe zum Beispiel oder mit echter Ananas, Mango usw. -------- Weitergeleitete Nachricht -------- Betreff: Fwd: Fwd: 26. Datum: Fri, 05 Dec 2014 21:16:09 +0100 Von: Markus <universe at truemetal.org> An: universe at truemetal.org Moritz - wie viele Spiesse / Zubehoer fuer wie viel Personen? -------- Weitergeleitete Nachricht -------- Betreff: Fwd: 26. Datum: Fri, 05 Dec 2014 21:12:04 +0100 Von: Markus <universe at truemetal.org> An: universe at truemetal.org Anna: Fructoseallergie -------- Weitergeleitete Nachricht -------- Betreff: 26....

Hello Everyone. We have: Asterisk 1.4.21.2 zaptel-1.4.11 libpri-1.4.5 Sangoma A101D Connected to a PRI Cicso 7960G phones (About 30 of them) We have a problem with dropped calls that has going on for a long time. We get up to 5 dropped calls on a bad day. They all seem to be incoming calls. I have a recording of what my users report a dropped call sounds like right before it drops

Dear all, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) This is the unweighted fit, in the code of 'nls' one can see that 'nls' generates a vector

Dear All, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) Now for a weighted fit: fm2DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal),

...truhe zum Beispiel oder mit echter Ananas, Mango usw. > > > -------- Weitergeleitete Nachricht -------- > Betreff: Fwd: Fwd: 26. > Datum: Fri, 05 Dec 2014 21:16:09 +0100 > Von: Markus <universe at truemetal.org> > An: universe at truemetal.org > > Moritz - wie viele Spiesse / Zubehoer fuer wie viel Personen? > > > > -------- Weitergeleitete Nachricht -------- > Betreff: Fwd: 26. > Datum: Fri, 05 Dec 2014 21:12:04 +0100 > Von: Markus <universe at truemetal.org> > An: universe at truemetal.org > > Anna: Fructoseallergie > > >...

Hello Bernardo, --------- If I understood your problem this script solve your problem: q<-0.15 + c(-.1,0,.1) h<-10 + c(-.1,0,.1) 5*q*h [1] 2.475 7.500 12.625 --------- OK, this solves the simple example. But what if the example is not that simple. E.g. P = 5 * q/h Here, to get the maximum tolerances for P, we need to divide the maximum value for q by the minimum value for h, and