Is there an Upsert API call to create/ update a subscriber in 1 API call?
Is there an option to create and Opt-in a user in the same API call?
Access network share from within VBScript eg FileSystemObject How to align content of a div to the bottom If temperature makes particles vibrate faster, and movement is limited by the speed of light, then I would assume that temperature must be limited as well. Why is there no limit? Transforming positively skewed data with positive and negative values My data (see image below) is positively skewed and contains both positive and negative values. I'd like to transform it to achieve normality so I can apply a repeated measures ANOVA, but can't find a technique that works.  I've considered the following options:   Square-rooting: doesn't keep the sign Square-rooting then making the original negative values negative: makes my data bimodal Natural log transform: can't do this with negative values Add a constant then log transform: this arbitrarily leaves my lowest value as an outlier '' transform (Cox, 2005; also Whittaker et al. 2005): makes my data bimodal Outlier removal (2.5 SD): doesn't fix normality, but an ANOVA on this data has similar results to an ANOVA on the raw data (for better or worse). Perform the non-parametric equivalent (Friedman test) and compare the results: will do this today   My supervisor suggests I should stick with the outlier removal though the data is still visibly skewed, and 'see what the reviewers suggest' but I just wanted to check here in case there was anything I was missing.  Original data:    With outliers (2.5 SD) replaced with the value at that criterion. I later decided to just exclude the outliers, so imagine the highlight data points in each condition removed. Find unique rows in numpy.array I added a user to a group, but group permissions on files still have no effect I changed permissions of a file (chmod g+w testfile) and running ls -l testfile gives:  -rwxrwxr-x 1 user1 user1 0 2011-01-24 20:36 testfile   I then added a user to that group ("/etc/group" has user1:x:1000:user2 line), but am failing to edit that file as user2. Why is this so? Sum $\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$ A series is given as follows   $$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$  Can you give me hints to get started finding its value? Thanks. NewForm.aspx disappeared Why does the division algorithm work for converting between number bases? When to use Stateful session bean over Stateless session bean? I have two tables, movies and categories, and I get an ordered list by categoryID first and then by Name.  The movie table has three columns ID, Name and CategoryID. The category table has two columns ID and Name.  I tried something like the following, but it didn't work.  var movies = _db.Movies.OrderBy( m =&gt; { m.CategoryID, m.Name }) How can an int be cast to an enum in C#? Short brandable domain vs. long keyword domain Can't login using ssh as root I am trying to virtualize an old linux server, and the PtoV software requires a password login as root via ssh. However, the root login is being prevented with a "Permission Denied" error on the client end, and a "Failed Password for root" on the server end (when sshd is running in debug mode).  I HAVE set PermitRootLogin yes in /etc/ssh/sshd_config.  This is an old RedHat 7.3 server that has had hardening scripts like Bastille run on it years and years ago, and has many remnants of old OS's (like ipchains - no longer supported by the kernel) lying around. All to say I have little idea what has been altered from the original system, but there is a lot.   I can log in fine as root from the console I can su to root fine from within the console or ssh session I can ssh fine as a regular user.   My question is: What else could be preventing my root login on this server?  More information: /etc/pam.d/sshd   auth       required     pam_stack.so service=system-auth  auth       required     pam_nologin.so  account    required     pam_stack.so service=system-auth  password   required     pam_stack.so service=system-auth  session    required     pam_stack.so service=system-auth  session    required     pam_limits.so  session    optional     pam_console.so   /etc/pam.d/system-auth  auth        required      /lib/security/pam_env.so auth        sufficient    /lib/security/pam_unix.so likeauth nullok auth        required      /lib/security/pam_deny.so account     required      /lib/security/pam_unix.so password    required      /lib/security/pam_cracklib.so retry=3 type= password    sufficient    /lib/security/pam_unix.so nullok use_authtok md5 shadow password    required      /lib/security/pam_deny.so session     required      /lib/security/pam_limits.so session     required      /lib/security/pam_unix.so     Can anybody tell me if the hashing algorithm could be the culprit?  Keeping in mind that I can log in as an unprivileged user. What is the rough estimate of salary value for a taxpayer to pay AMT? For US federal tax, whether a taxpayer is subject to AMT is determined by a complicated way in Instructions for Form 1040, Line 45, which involves an incomprehensible "worksheet to see if you should fill in form 6251".  By simplification and approximation, is there a rule of thumb for an estimate of a salary value of a taxpayer, so that a salary above the value might cause the taxpayer to pay AMT with a big possibility, and a salary below the value might not with a big possibility?  Why do the following two sources seem to differ greatly in estimates of the boundary salary value:    says that  the possible boundary salary value for a single taxpayer  to pay AMT is $53,900.  says that      Taxpayers with small incomes fall within the AMT exemption limits.   That leaves the upper middle class to face the tax, especially those   with incomes between $200,000 and $500,000    Thanks. Convert integers to written numbers Is there an efficient method of converting an integer into the written numbers, for example:  string Written = IntegerToWritten(21);   would return "Twenty One".  Is there any way of doing this that doesn't involve a massive look-up table? How can I start up an application with a pre-defined window size and position? I'm wondering is there any way to achieve the affect of the Ctrl-Alt-Keypad shortcuts in Unity using terminal commands instead? I want a command that sets a gui window to half the size of the screen, either left or right aligned.  By way of background, I'm writing a script that runs after log in. It uses Zenity to ask whether or not I want to open my development environment (GVim and IPython side-by-side). I have been trying to achieve two equal-sized windows for these programmes by using set lines= columns= in my .gvimrc and c.IPythonWidget.width = and c.IPythonWidget.height = in my ipython_qtconsole_config.py. However, there are problems associated with this approach. a) Let $a&gt;0$ and the sequence $x_n$ fulfills $x_1&gt;0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?  We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.  Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that   $$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$ b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit. &quot;Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit.&quot; I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance! Prove $ \forall c \in \mathbb{R^{+}}, \forall B \in \mathbb{N}, \exists n \in \mathbb{N}, (n \geq B) \wedge (\sqrt{2n} > c \ln(n+2)) $? I tried this way, I only need to know if this is correct or if there are better ways to solve this:  $2^{1000}$ does not have a factor of $5$ obviously therefore we can assume  $$ 10^{m} &lt; 2^{1000} &lt; 10^{m+1}$$ for some  $m$   Assume $ k = 2^{1000}$, then take log on both sides $\log k = 1000  \log 2 \approx 301.02999 &gt; 301$  Therefore $2^{1000}$ has $302$ digits.