Is there any way to get a total of all questions on a given site that have been closed for each reason? E.g.,   123 duplicate 456 OT 789 too broad   Bonus points if OT is separated by its reasons.
Are there any stats about the percentage of close reasons used to close questions   Successfully (preferably, based on the number of closed questions) Unsuccessfully (alternatively, the total percentags of close votes cast by users)   ?   Is there any way to get this from the query explorer?
Probability mean is less than 5 given that poisson distribution states it is 6 What is $\sqrt{i}$? If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?     Is it used or considered often in mathematics? How is it notated? So before I reformatted (to get rid of Windows), I simply used this command to open documents by putting it at the end of the ~/.bashrc file:  alias commandless= 'gnome-open ./Documents/the-linux-command-line.pdf; gnome-open ./Documents/linux.odt'   But now everytime I start the terminal, it says:  bash: alias: gnome-open ./Documents/the-linux-command-line.pdf; gnome-open ./Documents/linux.odt: not found   before even typing anything in, and when I run the command nothing happens, not even error messages. If $f$ is continuous with $f(x) = f(2x),f(1) = 3$, then what is $ \int_{-1}^{1}f(f(x))\,dx$? If $f(x)$ is a continuous function such that $f(x) = f(2x)$ and $f(1) = 3\;,$ Then $\displaystyle \int_{-1}^{1}f(f(x))\,dx$   $\bf{My\; Try::}$ Here $-\infty &lt;x&lt;\infty$ and Given $f(x) = f(2x)$  So Using Recursively $$\displaystyle f(x) = f(2^1x)=f(2^2x)=..........=\lim_{n\rightarrow \infty}f\left(2^{n-1}x\right)$$   OR we can write it as $$f(x) = f\left(\frac{x}{2}\right)=f\left(\frac{x}{2^2}\right)=........=\lim_{n\rightarrow \infty}f\left(\frac{x}{2^{n-1}}\right)$$  Now How can i prove that $f(x)$ is a constant function.  Help me  Thanks $\sum k! = 1! +2! +3! + \cdots + n!$ ,is there a generic formula for this? I came across a question where I needed to find the sum of the factorials of the first $n$ numbers. So I was wondering if there is any generic formula for this? Like there is a generic formula for the series: $$ 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} $$ or $$ 1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + n^{2} = \frac{n(n+1)(2n + 1)}{6} $$  Is there is any formula for: $$ 1! +2! +3! + 4! + \cdots + n! $$ and $$ {1!}^2 +{2!}^2 +{3!}^2 + \cdots + {n!}^2 $$?  Thanks in advance. If not, is there any research on making this type of formula? Because I am interested. How to combine multiple commands in terminal? I have the following commands.  cd import zcat urls1.sql.gz | mysql -u root -p urls cd /var/www/project1/ nano 1.php   As of now I'm executing it one by one.  Is there a way to combine those commands in one line? How can I convert an integer into its verbal representation? Is there a library or a class/function that I can use to convert an integer to it's verbal representation?  Example input:     4,567,788`   Example output:     Four million, Five hundred sixty-seven thousand, seven hundred eighty-eight Tilde (~) File in Macintosh Directory Last night I discovered what looks like an unusual file named "~" (tilde) in the root of my "Macintosh HD" directory on my MacBook.    Finder reports the file is 2.2MB in size, of kind "TextEdit.app Document", and was created on 22nd January 2014 (so about a month ago).  I haven't noticed any operational issues with OS X or it's applications.   I haven't yet touched the file but would obviously like to remove it.  I am very conscious that '~' is a special alias to the user home directory and thus I would appreciate any recommendations in removing the file in a safe manner that isn't likely to nuke my home directory or screw up the operation of applications or OS X. I don't know whether to delete or rename and whether to use Finder or the command line.   It's entirely possible that I created the file by accident since I've recently been drafting and running various shell scripts that use the ~, user home alias.  MacBook Pro 15" (Retina, Mid 2012) Yosemite Given a $10×10$ grid with $9$ red blocks and $91$ white blocks, in each step we color one red block black and after that one white block red until there are $91$ black blocks and $9$ red blocks. Prove that there exits a step in which a black block shares an edge with a white block. I ran a few trials and couldn't get past the fifth step without adjacency. But I don't know how to build a proof. Another way to evaluate $\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}$? What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$ $$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$ $$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$ $$-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }$$ $$ -\left(\frac{\sin 2x}{2} +\sin x \right) +c $$ Is there any other way to do so ? Is it possible to do it by substitution ? Valid US visa on expired passport. Can I travel to the US? My old passport is in my maiden name and has already expired. My B1/B2 visa to travel to United States is still valid but is stamped in my old passport with my maiden name.   My new passport has my new name (old name got changed after marriage) but has details (like old passport number, father and mother's name) related to my old passport.    Will I be allowed to travel to US as long as I carry the my old as well as new passport?  Or Do I need to again get the visa stamped on my new passport? Why do gaps appear in my mesh when I animate it? As I attempt to complete a "Walk cycle" with my model, the mesh at the pants line separates from the skin as the model moves or as I rotate the controls. This happens in many places on my model and it reveals the model interior. It is a one piece mesh, though, so I don't understand why this happens. Is this caused by weight painting problems? How can I fix this? High-end compact camera with swivel screen and running on AA batteries I'm looking for a new compact camera that:   has a  is small enough to be mountable on a  and, runs on AA batteries. good flexibility   Requirement (3) stems from the fact that I may be hiking in the wild for weeks on an end, with no opportunity to change a Lithium-Ion battery. If another solution permits me to take thousands rather than hundreds photographs on full charge, I may consider that as well.  Additionally, it would be nice (but not essential) to have:   the ability to handle high dynamic range a viewfinder physical buttons — I live in a cold climate, and don't like to use touch-screens   The upper price range for compact models is still good for me, and I don't mind a pricerange of €400–€800 (but I don't mind cheaper either, if still of good quality).  I've been the happy user of several models in the , but now my PowerShot A650 IS has died. Starting from , it seems requirement (3) kills of many cameras: the , (might also fail (2)), the , the , the ... Although enough models meet requirements (1), (2), and (4), as well as the (non-essential) additional features I listed, it seems in the past 5 years, cameras have moved away from using AA batteries. What camera models still exist that meet my requirements?    Last photograph taken with my Canon PowerShot A650 IS, taking on Christmas Day 2012 in the Caldera de Taburiente, La Palma, Canary Islands, Spain. What can I do when getting "We are no longer accepting questions/answers from this account"? This is a follow-up to .     When I click a link to an SE site in the Android app, it gives me an "open with" pop-up that has two options:         Chrome   Stack Exchange         I can then choose to open SE posts like this "Just once" or "Always."      The problem is that when I set the SE app to "always," and I click "open in browser,"   Except, the app crashes when trying to open it in browser.  LogCat (just right after selecting "Open in browser")  01-22 18:03:36.354: E/AndroidRuntime(16064): FATAL EXCEPTION: main 01-22 18:03:36.354: E/AndroidRuntime(16064): Process: com.stackexchange.marvin, PID: 16064 01-22 18:03:36.354: E/AndroidRuntime(16064): java.lang.NullPointerException: Attempt to invoke interface method 'int java.util.List.size()' on a null object reference 01-22 18:03:36.354: E/AndroidRuntime(16064):    at com.stackexchange.marvin.utils.Helpers.getNonMarvinViewIntent(Helpers.java:452) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at com.stackexchange.marvin.ui.question.QuestionDialog$1.onClick(QuestionDialog.java:39) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at com.android.internal.app.AlertController$AlertParams$3.onItemClick(AlertController.java:1108) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.widget.AdapterView.performItemClick(AdapterView.java:310) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.widget.AbsListView.performItemClick(AbsListView.java:1145) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.widget.AbsListView$PerformClick.run(AbsListView.java:3066) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.widget.AbsListView$3.run(AbsListView.java:3903) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.os.Handler.handleCallback(Handler.java:739) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.os.Handler.dispatchMessage(Handler.java:95) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.os.Looper.loop(Looper.java:148) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at android.app.ActivityThread.main(ActivityThread.java:5417) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at java.lang.reflect.Method.invoke(Native Method) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:726) 01-22 18:03:36.354: E/AndroidRuntime(16064):    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:616)   I suspect it's related to , but not sure.  Specs:   Device model: Nexus 5 Android version: Android 6.0.1 SE app version: 1.0.82 (beta) Installed browser: Chrome 47.0.2526.83 (latest at current writing) Any examples of life forms on Earth that could have survived on Mars or Moon? Do we know of any living forms on Earth that could continue to live on Mars? Why does $\sqrt{n\sqrt{n\sqrt{n \ldots}}} = n$? Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the   nth x = √(1st x √ 2nd x ... √nth x);   Then  $$\text{the "infinith" } x = x$$  Example:   $$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$ Try it yourself, type calc in Google search, hit  then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).  I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.  However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually. How do I clear all user defined symbols? How long would the conference be for? I want to know if this is grammatically correct " How long would the conference be for?". Given a set $\mathcal{S}$ of size $n$, let a $k$-partition of $\mathcal{S}$ be a grouping into $k$ disjoint classes $$(S_1 ,S_2,...,S_k)$$   Where the $S_i$ do not necessarily contain the same number of elements.  Let $C(S_1, S_2, ... S_k)$ be the set of combinations which can be made by taking exactly one element of each class, (i.e. $C(S_1, S_2, ... S_k)$ is the Cartesian Product $S_1 \times S_2 \times ...\times S_k$). For a given $(n,k)$, what is the minimum number of k-partitions $N$ such that  $\bigcup\limits_{i=1}^{N} C(S_1^{(i)}, S_2^{(i)}, ...,S_3^{(i)}) = $ All Possible Combinations  Example:  for $n=4$ and $k=2$, $\mathcal{S} = \{1,2,3,4\}$ we can choose:  $$ S^{(1)} = \left\{\left\{ 1, 2\right\}, \left\{ 3, 4\right\}\right\}$$ $$ S^{(2)} = \left\{\left\{ 1, 3\right\}, \left\{ 2, 4\right\}\right\}$$  Then $$C_{1} = \left\{\left\{ 1, 3\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 2, 4\right\}\right\}$$ $$C_{2} = \left\{\left\{ 1, 2\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 3, 4\right\}\right\}$$  and $C_1 \cup C_2 =$ All possible combinations and $N=2$.