I want to be able to total an entire column. I want excel to give me the total sum of all numbers in the column and ignore the blanks and the #N/A.   How would I complete this?
I have a range and need to sum it using SUM function but the result is N/A if there is any N/A value. How can I make the SUM function to treat the N/A value as 0 value?  Please help!
Using apt-get to remove packages and dependencies I want wp_link_pages (mutli-page posts) to display the page numbers, the word "previous" before those numbers, and a "next" after those numbers.  It would look like this:  Prev 1, 2, 3, 4 Next  I'm attempting to do this without a plugin.  Here's what I've tried so far, but it isn't working, it is only displaying numbers.  &lt;?php wp_link_pages(array(     'before' =&gt; '&lt;span style="clear:both; display:block"&gt;Pages',      'after'  =&gt; '&lt;/span&gt;',     'next_or_number'=&gt;'number',     'previouspagelink' =&gt; 'previous',     'nextpagelink'=&gt; 'Next' )); ?&gt; I've googled and also searched the answers here but couldn't find my problem so here goes:  I know how to bevel a curve using another object and when I try it with a normal circle it's fine. But I'm trying to use this flat oval shape to bevel a larger curve of the same shape and it's not following the curve at the correct rotation. I've tried everything, rotating the bevel object, clicking random options in the object data menu.. I can't seem to budge it from looking how it is in the screenshot. In the screenshot you can see the curve I've failed to bevel correctly and also the bevel object in the bottom right.  Basically, I want the flat part of the oval to be above and below the curve, not on the sides as it is. I hope that makes sense. Anyone know what I need to do? $X$ and $Y$ are independent $U(0,1)$ random variables.   I have to find $E[X|X&gt;Y]$.   I found it out by integrating $\int_0^1 \int_y^1 x dxdy$ to get $\frac{1}{3}$ as the answer but this is wrong according to the answer key. What is the mistake that I have made? How do I remove the LAMP stack so I can start over? I'd like to start over with my webserver and I'd like to reinstall all the LAMP packages, how do I do that? I'm trying to find the infinite sum that is defined by:  $$ 3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots $$  However, I do not know of any known formula to do this. Am I missing something really simple? Thanks! I'm running a VPS of a german provider and want to switch from Ubuntu to CentOS. There I tried to install firewalld (was not preinstalled), but it always fails with the error  ERROR: Failed to load nf_conntrack module: modprobe: ERROR: could not find module by name='nf_conntrack' modprobe: ERROR: could not insert 'nf_conntrack': Function not implemented modprobe: ERROR: Error running install command for nf_conntrack modprobe: ERROR: could not insert 'nf_conntrack': Operation not permitted   Obviously the nf_conntrack kernel module was not found. Since this is a VPS, I cannot tinker with the kernel modules. On the support site of my provider I found a list of available kernel modules, for example:  ip_conntrack_netbios_ns ipt_conntrack    ip_conntrack    ip_conntrack_ftp    ip_conntrack_irc   Since nf_conntrack is not listed here, does that mean, that I cannot use firewalld? Is it possible to run it without this module?  I guess I could use iptables, if I wanted to (it seems to run correctly). But as I have more experience with firewalld, I would like to stick with it.    CentOS Linux release 7.7.1908 (Core)  uname -a: Linux xxxxxx.xxxxxxxx.xxx 3.10.0-042stab140.1 #1 SMP Thu Aug 15 13:32:22 MSK 2019 x86_64 x86_64 x86_64 GNU/Linux Show that $(l_1)^* \cong l_{\infty}$ Nautilus 3.6 doesn't have a status bar? In 12.04, Nautilus showed the amount of remaining space in the status bar , but as of 13.04 the status bar is gone.  gsettings set org.gnome.nautilus.window-state start-with-status-bar true   turned out "no such key".  Is there any way to bring back the status bar? For any triangle with sides $a$, $b$, $c$, prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$   This is IMO 1983 problem 6.  I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except wasting 3 pages that lead to nothing (please don't mind the joke). Using $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ also didn't lead to anything for me. Could you give me a hint for finding the proper substitution? I was wondering about the reason of the choose of the X as symbol used to represents the "Close application" operation. General formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^ae^{-bx}dx$ Find a general formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^a e^{-bx}dx.$   I've been pondering about this question for some time. Obviously integration by parts is way too inefficient. I know there's recursion involved and maybe there's a way to find a general formula through intuition. Prevent Photos app on OSX from launching when plugging in device The Photos app opens whenever I plug in my phone.    How can this be prevented?  I have seen many solutions for the old iPhoto app that involve changing the "Connecting camera opens" option, but the new app has no such option. Eigenvalues of symmetric matrices are real without (!) complex numbers Is there any proof of the fact that the eigenvalues of symmetric matrices (i.e. $A\in\mathbb{R}^{n\times n}$ with $A^t=A$) are real without the use of the concept of complex numbers? Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this? Obtaining knowledge from a random forest Random forests are considered to be black boxes, but recently I was thinking what knowledge can be obtained from a random forest?  The most obvious thing is the importance of the variables, in the simplest variant it can be done just by calculating the number of occurrences of a variable. The second thing I was thinking about are interactions. I think that if the number of trees is sufficiently large then the number of occurrences of pairs of variables can be tested (something like chi square independence).  The third thing are nonlinearities of variables. My first idea was just to look at a chart of a variable  Vs score, but I'm not sure yet whether it makes any sense.    Added 23.01.2012 Motivation    I want to use this knowledge to improve a logit model. I think (or at least I hope) that it is possible to find interactions and nonlinearities that were overlooked. "All sites" overview should not require scrolling when selecting a site I thought that the quotation mark (") was simply a type of grouping marker but I'm debugging some NHibernate code and notice that while  SELECT * FROM site WHERE site_id = 3;   Works fine  SELECT * FROM "site" WHERE site_id = 3;   fails with a table or view does not exist error.  What gives? How is Cooper's age justified? Run a batch file in a completely hidden way I'm looking for some way to run a batch file (.bat) without anything visible to the user (no window, no taskbar name, .etc..).  I don't want to use some program to do that, I'm looking for something cleaner. I've found  that uses VBScript, but I don't really like using VBS, either.