Yesterday I  and after 2 hours I found a "solution", I would answer, for the first time, at my question and Stack Overflow block my answer with this message:     Oops! Your answer couldn't be submitted because:           Users with less than 100 reputation can't answer their own question for 8 hours after asking. You may self-answer in 5 hours. Until then please use comments, or edit your question instead.)      and I would want to understand why there is this block?  Yesterday I worked around it by adding on my question a block [answer]answer[/answer] as shown at screen shot below and today I modified again the post moving the block out into an answer but, for me, it is strange to have self-answer block.  EDIT: because Cody and Rory ask why I add the tag [ANSWER] on my question the answer is simple: I didn't know how tell that I resolved the problem so I create that tag for underline that it was out of my question and temporarly there...
If we're implementing , why don't we also require that users have at least 1.5 times the amount of reputation needed to ask their current question in order to answer their current question?  I believe this would really reduce the amount of non-answers that should have been edits to the original question, or perhaps a comment under another answer.  Edit  Inspired by marcog's excellent suggestion , perhaps it would make more sense to set a minimum amount of time that must pass before you can answer your own question. New users would need to wait in the neighborhood of six hours, while users with more reputation could answer their own question within minutes if they discovered the solution on their own.  This would get in the way of the behavior that we don't want, which is people posting answers that should have been edits while not annoying most.
Can't finish registration in Stackoverflow.ru I have an account on Stack Overflow and also on Russian Stack Overflow. I can ask questions on Russian Stack Overflow, but it always displays the phrase "Finish registration". I have tried logging in by different methods — Facebook, Google, etc. — but it still says that my registration is not finished. Can someone help me with it? or explain what I should do there to finish the registration?  P.S: I can't write questions in Meta Russian Stack Overflow because it says that my registration is not finished.  P.S.S: I have already tried "Click the "register" link at the top of the page." it brings to the login page with the phrase "Finish signing up by using any of the following options:". I have already tried all options there. I also can't merge them because they already linked. If I live in the US, for example, where current interest rates are practically nil, how can I take advantage of the fact that ? How to interpret regression coefficients when response was transformed by the 4th root? I'm using fourth root (1/4) power transformation on my response variable, as a result of heteroscedasticity. But now I'm not sure how to interpret my regression coefficients.   I assume that I would need to take the coefficients to the fourth power when I back-transform (see below regression output). All of the variables are in units of dollar in millions, but I would like to know the change in dollar in billions.    While holding the other independent variable constant, a billion dollar change in fees, on average, leads to a change of 32 (or 32,000 dollars) in collections. I take 0.000075223 * 1000 (to get to billions) ^ 4 = 0.000032. Now do I multiply this number by 1 million or 1 billion (the original unit of the dependent variable is in millions)?  lm(formula = (Collections^(1/4)) ~ Fees + DIR)                   Estimate      Std. Error  t value            Pr(&gt;|t|) (Intercept)   2.094573355     0.112292375   18.653  0.0000000000000151 Fees        **0.000075223   **0.000008411    8.943  0.0000000131878713 DIR           0.000022279     0.000004107    5.425  0.0000221138881913 Given $S \hookrightarrow T$ construct $U ≈ T$ disjoint from $S$ in Z set theory? Lattices are congruence-distributive find and echo file names only with pattern found I was viewing  question in the Android app. As shown in the below screenshot, while the top shows "4 answers", the bottom button will ask to "View 6 Answers". After checking the question, there are indeed six answers to the aforementioned question.    I am using Android 4.3 with the 0.1.64 version of the app.    EDIT: The changes to the UI made this question obsolete. Why is every Noetherian zero-dimensional scheme finite discrete? Which is more correct:     Apropos of your earlier comment, I decided to....   or      Apropos your earlier comment, I decided to...   Actually, apropos is so fancy a word, even I, a word maven if I do say so myself, am not entirely sure when to use it.  If I may add to my preposition question, are there any constructions where it would be appropriate to use something like with regard to and not apropos? When using sudo to allow edits to files, I regularly get 'permission denied'.  For example, my mouse is jittery and sluggish, so I want to disable polling:  sudo echo "options drm_kms_helper poll=N"&gt;/etc/modprobe.d/local.conf   I'm prompted for a password, and then get:  bash: /etc/modprobe.d/local.conf: Permission denied   So I tried to do a temporary change to disable polling by using:  sudo echo N&gt; /sys/module/drm_kms_helper/parameters/poll   Yet again the system responded with:  bash: /sys/module/drm_kms_helper/parameters/poll: Permission denied   Any ideas? $A_n:Y\rightarrow Z$ are operators that strongly converge to $A$. Also, $\|A_n\|_\text {op}\le c$ for $c&gt;0$. Given a compact operator $T:X\rightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).     I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas? I'm struggling to understand the concept of quantum entanglement. I've distilled my understanding into an analogy, and I need your help to validate it. Here it is:  Let's say I receive two envelopes. Both envelopes have this written on them:     Hello! Open this letter...   If the paper inside is red, then other envelope has blue paper.   If the paper inside is blue, then other envelope has red paper.   So, if I open one letter, and the paper inside is red, then I know the other letter contains the blue paper.  Is this equivalent to quantum entanglement? The fact that I will know the color of the other paper when I open (i.e. observe) one envelope?  In theory, if one of the letters "appeared" in a different galaxy, I would still know the color of the paper inside that letter, just by opening my letter, instantly, correct?  Before I open the letter, is the paper inside each envelope in superposition (??), i.e. 50/50, red or blue?  In this analogy, I am assuming that what's written on the letters is true. Is this a correct assumption in quantum physics? I think so yes?  In real quantum physics, would I be able to change the color of one paper, so that I change the color of the other one? No right? Short story about a crew finding an ancient alien ship filled with tons of artifacts from their society I'm wondering if there's any log which stores information about a particular program such as when the program has started and closed manually on Windows such as MS Word, or Chrome. Any ideas? GRUB and LILO both fail to install to NVMe hard disk when installing Debian Has the wall been repaired? If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this? Disable chat migration notification if one of the users has insufficient rep? How to make a bib entry appear also in the main document Why does boiling water in the microwave make a cup of tea go weird? When I boil water in the kettle, it makes a nice cup of tea.  Sometimes I need to use a microwave because a kettle isn't available.  I boil the water in the mug and it looks pretty normal, but when I drop in the teabag the water froths up and looks foamy.  I don't see what the chemical difference is here, so I assume it must be some physical difference.  I have noticed this with multiple types of tea and multiple microwaves, the results being consistent so it's not just a weird microwave or something like that.    What is the reaction here and how/why does it occur?  Here is a photo of the 'fizzy' looking tea just after dunking in the teabag.