My flag limit is increasing day by day. I haven't noticed before how many flag I can raise. But since I am noticing it is increased a lot. In last week, I have raised above 200 helpful flags.   I have noticed to increase it from 36. Daily it is increasing by 1. Now it is reached to 40.  Is it good/ bad? On what basis, flag limit is getting increasing daily?
Looks like flagging has been affected by releases today. Are these intended effects?   Higher daily limits (appear doubled for me, flagging of posts limit also doubled).  Explained by Shog9 below. Soft comment deletion  - e.g. for "accept rate" obsolete comments - it does in fact work, but the immediate feedback is gone.  This is fixed now. Has the throttling rate changed?  Seems possible to repeated flag without waiting 5 seconds.  If it was broken, it isn't anymore. Old soft deletions that were neither categorised as "helpful" or "declined" have all become "helpful".   If limits have been updated, can we get the new formula published? Not changed, as confirmed by Shog9.
At a movie theater, the whimsical manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket.  You have the option of choosing any position in the line. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout the year (365 days year), what position in line gives you the best chance of getting free ticket? How coherent light is created for single slit experiment? Prove that if $n$ is a composite, then $2^n-1$ is composite. Not sure if I'm doing this correctly but this is what I've done:  Assume that $n$ is composite and suppose $2^n-1$ is a prime for $n \gt 2$. Then, $2^n-1 = 2k$ for some $k \in \Bbb Z $, $\forall n$.  But this would be a contradiction since $n \gt 2$.  I'm not sure if this a correct proof.  EDIT: I realized that it's to prove $2^n - 1$ NOT $2^{n-1}$ so my attempt above is completely wrong. Using proof by contradiction vs proof of the contrapositive Uploading video to youtube in right format I have a Sony CX115 and I have a little footage I'd like to put on youtube. I've downloaded Arista from the Ubuntu Software Center and then I've downloaded its Youtube HD plugin. I've converted my video and uploaded it to youtube... The result is a flashy video with a lot of grey in it:    What could I do to publish it in high quality so it'll be viewable? What does the Y capacitor in a SMPS do? How do I install the Nvidia drivers? I just ordered the Nvidia GTX card. I have a dilemma, though. Should I keep using the driver which is available in "additional drivers" in Ubuntu, or should I install the driver from the Nvidia site?  So which driver is the best for me? Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence Given  $$ y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1. $$ Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ?  I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck. create a .deb Package from scripts or binaries I searched for a simple way to create .deb Packages for things which have no source code to compile (configs, shellscripts, proprietary software). This was quite a problem because most of the package tutorials are assuming you have a source tarball you want to compile. Then I've found  short tutorial (german).  Afterwards, I created a small script to create a simple repository. Like this:  rm /export/my-repository/repository/* cd /home/tdeutsch/deb-pkg for i in $(ls | grep my); do dpkg -b ./$i /export/my-repository/repository/$i.deb; done cd /export/avanon-repository/repository gpg --armor --export "My Package Signing Key" &gt; PublicKey apt-ftparchive packages ./ | gzip &gt; Packages.gz apt-ftparchive packages ./ &gt; Packages apt-ftparchive release ./ &gt; /tmp/Release.tmp; mv /tmp/Release.tmp Release gpg --output Release.gpg -ba Release   I added the key to the apt keyring and included the source like this:  deb http://my.default.com/my-repository/ ./   It looks like the repo itself is working well (I ran into some problems, to fix them I needed to add the Packages twice and make the temp-file workaround for the Release file). I also put some downloaded .deb into the repo, it looks like they are also working without problems. But my self created packages didn't... Wenn i do sudo apt-get update, they are causing errors like this:  E: Problem parsing dependency Depends E: Error occurred while processing my-printerconf (NewVersion2) E: Problem with MergeList /var/lib/apt/lists/my.default.com_my-repository_._Packages E: The package lists or status file could not be parsed or opened.   Has anyone an idea what I did wrong?  UPDATE 2012-03-06: Just a a hint for another person who is searching for a easy way to create DEBs: Take a look at . What can I do with .chk files in FOUND.000 folders? Just discovered I have gotten a FOUND.000 folder on my usb drive. I guess from running chkdsk once. It has three files in it:   FILE0000.CHK FILE0001.CHK FILE0002.CHK   What exactly can I do with these files? What are my options? Are they lost files? Are they garbage? Can I use them for anything at all? Or will I just have to delete them? Inverse metric in linearised gravity From what I've read, in the framework of linearised gravity, one perturbs the metric around a Minkowski background, $\eta_{\mu\nu}$, such that  $$g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x)\tag{1}$$  where $h_{\mu\nu}(x)$ is a small perturbation, i.e. $$\big\lvert h_{\mu\nu}\big\rvert&lt;&lt;1.\tag{2}$$  The inverse metric is then found by assuming the following ansatz:  $$g^{\mu\nu}(x)=\eta^{\mu\nu}+\tilde{h}^{\mu\nu}\tag{3}$$  where $\tilde{h}^{\mu\nu}$ is also small (i.e. $\big\lvert \tilde{h}_{\mu\nu}\big\rvert&lt;&lt;1$).   Using this, it is easy to find that  $$g^{\mu\nu}(x)=\eta^{\mu\nu}-h^{\mu\nu} \tag{4}$$  to first order.   My question is, what is the justification for this ansatz? Is it simply that one expects the inverse metric to have a similar form to the metric in order to satisfy $$g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}_{\;\nu}~?\tag{5}$$ I'm pretty new to this and this is my first question here, any help would be greatly appreciated.  So basically what I want to achieve is a floor material as seen in this photo...    ...the reflections are very crisp but dim.  With a glossy shader I get the following results:    First picture: reflections are blurry  Second picture: roughness of the material looks right but reflection is bright red  Third picture: If tinted in any other color than white, it seems to produce the result I want. Reflection is sharp and gets tinted blue. What I need is the reflection to be tinted white.  Thanks for any help. In linear regression, each predicted value is assumed to have been picked from a normal distribution of possible values. See below.  But why is each predicted value assumed to have come from a normal distribution? How does linear regression use this assumption? What if possible values are not normally distributed? In mathematics, we like talking about idempotent matrices. In a masters program that I am currently enrolled in, I have heard one professor pronounce it one way, and another professor pronounce it another way. The  explains it better than I could:  Is there a correct pronunciation for idempotent? Is it like omnipotent (om-nip-o-tent), so id-EMP-o-tent, or is it more like the two seperate words idem + potent I'd say EYE-dm-POT-nt, but then I speak Brit. Charles Matthews 07:57, 30 Jun 2004 (UTC) Ditto, but then I speak Australian ;) Dysprosia 08:01, 30 Jun 2004 (UTC) Actually the British is more like EYE-dm-PO-tnt, I guess. Charles Matthews I'd just like to pedantically point out that it would be pronounced i-DEM-po-tent because because syllable onsets are maximized. daesotho 20:33, 21 Oct 2004 (UTC) The American pronunciation puts primary stress on the second syllable ('dem') and secondary stress on the fourth syllable ('tence'). Could somebody more familiar with IPA than I am please add this to the article as an accepted pronunciation? My American dictionary says |ˈīdemˌpōtənt| for the adjective. I speak American (non-natively) and I wouldn't stress the second syllable for itempotence. My only question is, right now it says /ˌaɪdɪmˈpoʊtəns/ -- but can't it be stressed on the first syllable too? Should we add /'aɪdɪmˌpoʊtəns/ as an alternative? -- 87.160.141.177 (talk) 14:50, 3 June 2011 (UTC) And by the way I think it's silly giving a pronounciation guide as this is just imposing a particular accent. (For what it's worth I pronounce idempotent with a short 'i', as in the word 'id'). Alex Selby I agree with Alex Selby - the pronunciation guide should be extended to show the variation, or removed. Strictly, there's no reason for the first i to be long (neuter idem has a short first syllable in Latin), and I can't find any authoritative source which advocates the pronunciation given. Ms7821 (talk) 22:59, 15 July 2015 (UTC)  Is there a &quot;correct&quot; pronunciation for this word? I know how to monitor the network interfaces on my local computer with Wireshark. But, how do I monitor the traffic flowing through my wireless access point from connected computers?  It seems like this might not be possible without some kind of special software on the wireless access point itself. Are there other schools for witches and wizards in the rest of the world? In the books they only speak of 3 schools, but what about Asian or American people?  And even more, were they affected by the events happening in Europe? How do I open the user folder of a hard drive that I transfered? My Windows 10 notebook died but fortunately I was able to rescue the hard drive and put it into another PC.   Unfortunately, the other PC can't open the C/Users/Joe folder. What do I have to do to open the folder? Reflexive and relation Please give me feedback on my answer to this question. Question:  For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;  $x\equiv y$   if $x^{2}=y^{2}$  on $R$  To show $\equiv$   is reflexive, we need to show that ,  $\forall x\in R:X\equiv x.$  Let $x\in R$,   $X\equiv x$   if $X^{2}=x^{2}$  , which is obvious.  $[x]\triangleq:\{y\in R/x\equiv y\}$  $[0]=[y\in R/0\equiv y\},$  when $y=0$,  Then $y^{2}=0^{2}=0$  Thus; there are not many infinity many equivalence classes. Currently I have some code that is being generated dynamically.  In other words, a C# .cs file is created dynamically by the program, and the intention is to include this C# file in another project.  The challenge is that I would like to generate a .DLL file instead of generating a C# .cs file so that it could be referenced by any kind of .NET application (not only C#) and therefore be more useful. How to attach my repo to heroku app