What does Axiom of Choice mean
Can you explain the "Axiom of choice" in simple terms?
Why is my membership length only "Six months"? I was viewing the  and I noticed that I've apparently only had my account for six months, which is news to me... I opened my first site account here in February of 2014.  My account is much older than this, what's causing this bug?  To add some detail... On the per-site list, it does line up the &quot;member for X&quot; text with my actual membership time on that site. For example, when I look at the Meta Stack Exchange leaderboard, it shows me as a member for 2 years, 10 months. This is expected.  Update. With some work, today, I've managed to exceed my Interpersonal Skills hat count of with my Meta Stack Exchange hat count. This has changed my account age to match my MSE account age...  Still not reflective of my actual age... but better. Motion Tracking objects that go out of frame and re-enter later I want to know, if this bug is fixed, because all previous versions of Ubuntu didn't work for me. So I just want to know, should i leave Windows 10 and go for Ubuntu 17.10. Thanks! Finding cartesian product with PHP associative arrays Should I use 'or' or 'nor' after a negative statement? In the last episode of season 7, The Doctor goes into his timeline to save Clara. It ends when they see his war regeneration and Clara faints. Did I miss a special episode between that one and the 50th anniversary special? Because they never explained how they escaped his timeline. How do I find the location of Python module sources? How do I learn where the source file for a given Python module is installed? Is the method different on Windows than on Linux?  I'm trying to look for the source of the datetime module in particular, but I'm interested in a more general answer as well. An arithmetic function is defined as follows:$f(1)=1$,  $f(2k)=k$ and $f(2k+1)=f(k)+f(k+1)$. When (for which $k$) is $f(k)$ even?  While it is obvious that $f(4n-1)=f(4n)=2n$, therefore $f(k)$ is even for $k=4n$ and $k=4n-1$, which is possible to prove using induction (isn't it?), I don't know how to prove the converse.  I tried to solve the recurrence or make use of constant difference between certain elements of the sequence $\ldots$ What am I missing?  Thank you. What is role-playing, and where do I start? I have been told that roleplaying is great fun. I've looked and looked and it all seems so complicated. Could anyone tell me a good place to start, and what sort of things you end up doing in roleplaying? What is projection of WGS84 in GIS software? WGS84 is a reference ellipsoid that determines coordinates of every point on Earth using latitude, longitude and height about the its surface.  This ellipsoid is often used as a base to make maps. Though, to make a map on a flat paper or a flat digital screen you need a projection too. An example of a popular projection would be UTM.   Mainstream GIS software give plethora of projection alternatives to display your data layers. Among them, they also let you display your data using WGS84 only.  I think they are using a certain projection but what projection is that? Is $(\mathbb{R}^2, lexicographic) \cong (\mathbb{R}, discrete)\times (\mathbb{R}, usual)$? I have the following question:  Is $(\mathbb{R}^2, lexicographic) \cong (\mathbb{R}, discrete)\times (\mathbb{R}, usual)$?  Note: the symbol $\cong$ denotes a homeomorphism between topological spaces  What I think:  I think that in fact there is a homeomorphism between those spaces, because is possible to express the lexicographic order as union of vertical lines with open extremes, I mean this is a basis for the lexicographic order. If I consider the Identical function I have the homeomorphism. Am I right? Not sure if this option ever existed, but the new QGIS version hides a lot of things somewhere else and this seems so simple that I'm kind of convinced it was possible to do this before. I have quite a large data set with ship tracks. I now want to analyse individual ship tracks (made from point layer with  Plugin). However, it is so crowded that it is hard to look at individual tracks. I don't want to save each single track as a new layer, so I wonder if it is possible to select one track in the Attribute table and display only that one on the map/hide the others? Why is my query suddenly slower than it was yesterday? [Salutations]  (check one)  [ ] Well trained professional, [ ] Casual reader, [ ] Hapless wanderer,   I have a (check all that apply)  [ ] query [ ] stored procedure [ ] database thing maybe     that was running fine (if applicable)   [ ] yesterday [ ] in recent memory [ ] at some point    but is suddenly slower now.  I've already checked to make sure it's not being blocked, and that it's not the victim of some long running maintenance task, report, or other out of band process.  What is the problem, what should I do, and what information can I provide to get some help?  [*Insert appropriate closing remarks*] How can I assign the output of a command to a shell variable? Intuition for why the sum of Bernoulli trials is a sufficient statistic via cancellations on numerator and denominator? Suppose that $X_1, \ldots, X_n$ are iid Bernoulli trials with parameter $p$, and that $Y = \sum_{i=1}^nX_i$. Then, we have that $Y$ is a sufficient statistic by the following condensed proof:  $$ P(X_1 = x_1,  ... , X_n = x_n |Y = y) = \frac{p^y(1-p)^{n-y}}{\binom{n}{y}  p^y(1-p)^{n-y}} =\frac{1}{\binom{n}{y}}   \text{ if } \sum_{i=1}^{n}x_i = y $$  I am trying to see why this is the case. My reasoning is that there is some cancellation property on top and bottom that "kills" off the parameter $p$. In that whatever is on top through $p$ is perfectly matched by whatever is on the bottom minus scaling factors (in this case $\binom{n}{y}^{-1}$).   However, I am wondering what the deeper meaning behind this is, that the numerator of the joint distribution of the data and the sufficient statistic has its EXACT form on the bottom, except for constants which are only multiplicative (to facilitate cancellations). Am I onto something bigger and deeper here? Can we implement a dead time on answers for high reputation user Summary: High reputation users on SO come swooping in the second a simple question is posted, and get the first answer in before I've even opened my Python interpreter to test my idea out.  This actually is a problem because it prevents use lower users who aren't as good  at programming from getting any answers in.  This makes it very hard to build up reputation.  Thus, I propose a short dead time for high reputation users where their answers do not appear.  The Problem: This isn't just reputation.  By answering questions and having discussions with OP, we all get better at our language.  Teaching is one of the best ways of learning, and a more well versed population is absolutely good for SO.  (If anyone wants to argue that, go home)  While its nice to have folks who can quickly answer nearly all questions in a jiffy, its a little frustrating for those of use who simply aren't as fast, but still want to contribute.  The result is that many people do not answer questions because they feel like they just can't get an edge in.  SO should aim to incentivize as many answers by as many users as possible.  The most answers from the most users is the most beneficial situation for the site and the community. Once we accept this idea, it is clear that something needs to be done to address the issue at hand    The Solution: Luckily, this is an easy problem to solve; we use a dead time.  High reputation users can post answers whenever they like, but they won't appear until after the dead time is completed.  The trick is balance, a dead time which is too long will harm the OP as the question will remain unanswered, however, a dead time which is too short won't actually fix the problem.  There definitely needs to be a discussion about what "high reputation" means and what dead time to use, but it should be independent of whether y'all agree that this is a problem.  So, what do you think?  EDIT: Wow, -18, I leave for lunch for 30 minutes and this is what happens!  I guess y'all think this is a bad idea, but let me try to clarify a few things on which I was unclear earlier.  This is not about reputation.  This is not about getting the first answer just for the sake of getting to say "in before..."  This is about learning.  By getting a shot at answering questions before the high reputation users, people learn. Think of it as doing an example problem in class before the teacher does it. You get a chance to try it yourself, test your knowledge, and if you fail, then the teacher is right behind you to show the correct solution.  This is what I'm trying to accomplish. However, seeing the solution, and then trying to answer the question, is not very helpful. This is how I view the status quo.    In response to Mr. Pieters' answer and comments: You're absolutely right, this is not about quantity, its about quality.  The fastest answer isn't always the best.  But asking the high reputation users to take a step back for a few minutes is a means of getting there.    Hopefully this clears things up a bit. Can I travel within the Schengen area if I currently don't have a valid visa or residency permit? How do you calculate the probability density function of the maximum of a sample of IID uniform random variables? Given the random variable  $$Y = \max(X_1, X_2, \ldots, X_n)$$  where $X_i$ are IID uniform variables, how do I calculate the PDF of $Y$? Prove that, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. JAVA Bad variable reference? Receiving strange output - possibly memory address?