The Room of Requirement is extremely useful, but Harry barely uses it. Sure the DA uses it, and so does Harry to hide his Potions book, but for half of the fifth book, and the sixth, whenever he needed anything (minus food) the room would bring it to them. So if he was worried about his O.W.Ls, then he could tell the room to help him out. Or if he wanted a secret passage to Grimmauld Place, he could use the room, instead of sneaking into Umbridge's office.   This is not asking what the room's limits are, but asking why Harry doesn't frequently use it.
Can the Room of Requirement provide anything if needed? For example:   James Potter wanted to hide from Voldemort, it would have served as a perfect hiding place. Harry Potter wanted all the Hocruxes, he should have gone straight to the room of requirement where he would have found all the Hocruxes along with the Sword. All the muggle borns of the wizarding community could have taken refuge in the room of requirement. Can someone meet a dead person in the room of Requirement if needed? Would it have been possible for Harry to go in past in order to see Dumbledore for the explanation of many things? Can the room duplicate something? For instance, Harry needed the Elder wand; with Voldemort having it, could he still get it? Someone wishes to be knowledgeable than Dumbledore, would the seeker find his desired wish by finding books and things?   Who created the room of requirement (surely not Salazar)? Was it beyond Dumbledore's magic to create such a place?
The formula for a one sample t-test, as given by  is  $t = \dfrac{\bar{x} - \mu_0}{s\big/\sqrt{n}}$  However, in this equation, if the sample standard deviation, $s$, is zero, $t$ will be undefined. Is there any reasonable way to get a $t$ value in this case? Is it possible to run Microsoft Office 2010? I'm running the Storm King's Thunder campaign, and we are at the portion where the PCs get an airship to travel around in, the airship is about 1000ft in the air. There are approximately 5-6 lvl 7 PCs, and they have 3 or 4 hard encounters a day. There are two wizards, a warlock, ranger, paladin, blood hunter, sorcerer, and fighter. For the most part, they have had fun with the different weapons and enemies they have faced; however, I find that one or two spellcasting PCs have been using spells such as mind sliver, polymorph, and hypnotic pattern to defeat more challenging encounters, such as one with a roc and wyvern. They make the creature make saving throws with hypnotic pattern and cause the enemy to fall to its death, and using mind sliver makes it even harder, especially on low wisdom creatures like rocs. Or with the polymorph spell, they cause the enemy to become a fish or chicken and throw it off the edge, and since the damage carries over, it kills the monster. I realize it was legal and made fights funnier, but it does get annoying when it's happening in almost every fight. Are there any suggestions on how to keep them on track without them making every encounter easy? I thought of giving the monsters immunity to these effects, but it just sounds like a bad excuse for me not being prepared. A comparison between the verbs (come in / by), (drop in / by), (check in / by) (step in / by) and come over scp copy over ssh doesn't work - permission denied error, please? Multivariate linear regression vs. several univariate regression models In the univariate regression settings, we try to model   $$y = X\beta +noise$$  where $y \in \mathbb{R}^n$ a vector of $n$ observations and $X \in \mathbb{R}^{n \times m}$ the design matrix with $m$ predictors. The solution is $\beta_0 = (X^TX)^{-1}Xy$.  In the multivariate regression settings, we try to model   $$Y = X\beta +noise$$  where $y \in \mathbb{R}^{n \times p}$ is a matrix of $n$ observations and $p$ different latent variables. The solution is $\beta_0 = (X^TX)^{-1}XY$.  My question is how is that different than performing $p$ different univariate linear regression? I read  that in the latter case we take into consideration correlation between the dependent variables, but I don't see it from the math. How to convince longtable to do a page break earlier using the longtable package to typeset page spanning tables, I run into a nasty behaviour: only a dummy table head and the caption are put on the following page. The contents of the table is extracted from program sources. I don't have much influence here.  \documentclass[a4paper,twoside,DIV10,BCOR1.5cm]{scrbook} \usepackage{longtable} \newlength{\tablcol} \setlength{\tablcol}{4.7cm} \newlength{\tabrcol} \setlength{\tabrcol}{8.1cm} \itemsep0ex plus0.2ex % \begin{document} \begin{longtable}{p{\tablcol}p{\tabrcol}}% % \hline \multicolumn{1}{l}{{\bfseries Feld}} &amp; \multicolumn{1}{l}{{\bfseries Beschreibung}}\\ \hline \endfirsthead % \hline \multicolumn{2}{l}% {{\small\slshape Fortsetzung der vorherigen Seite}}\\ \hline \multicolumn{1}{l}{{\bfseries Feld}} &amp; \multicolumn{1}{l}{{\bfseries Beschreibung}}\\ \hline \endhead % \hline \multicolumn{2}{l}% {{\small\slshape Fortsetzung auf der n\"achsten Seite}}\\ \hline \endfoot % \hline \caption[show user output]{Beschreibung der Output-Struktur des show user Statements} \endlastfoot % \label{tucshowxuser}% {\scshape id} &amp; Die Nummer des Repository Objektes \\ {\scshape name} &amp; Der Name des Objektes \\ {\scshape creator} &amp; Name des Benutzers der dieses Objekt angelegt hat \\ {\scshape create\_time} &amp; Datum und Uhrzeit der Erstellung \\ {\scshape changer} &amp; Name des Benutzers der dieses Objekt zuletzt ge\"andert hat \\ {\scshape change\_time} &amp; Datum und Uhrzeit der letzten \"Anderung \\ {\scshape privs} &amp; Zeichenkette die K\"urzel f\"ur die Benutzerrechte auf dieses Objekt enth\"alt \\ {\scshape commenttype} &amp; Typ des Kommentars \\ {\scshape comment} &amp; Kommentar zum Objekt, wenn vorhanden \\ {\scshape id} &amp; Die Nummer des Repository Objektes \\ {\scshape name} &amp; Der Name des Objektes \\ {\scshape creator} &amp; Name des Benutzers der dieses Objekt angelegt hat \\ {\scshape create\_time} &amp; Datum und Uhrzeit der Erstellung \\ {\scshape changer} &amp; Name des Benutzers der dieses Objekt zuletzt ge\"andert hat \\ {\scshape change\_time} &amp; Datum und Uhrzeit der letzten \"Anderung \\ {\scshape privs} &amp; Zeichenkette die K\"urzel f\"ur die Benutzerrechte auf dieses Objekt enth\"alt \\ {\scshape commenttype} &amp; Typ des Kommentars \\ {\scshape comment} &amp; Kommentar zum Objekt, wenn vorhanden \\ {\scshape id} &amp; Die Nummer des Repository Objektes \\ {\scshape name} &amp; Der Name des Objektes \\ {\scshape creator} &amp; Name des Benutzers der dieses Objekt angelegt hat \\ {\scshape create\_time} &amp; Datum und Uhrzeit der Erstellung \\ {\scshape changer} &amp; Name des Benutzers der dieses Objekt zuletzt ge\"andert hat \\ {\scshape change\_time} &amp; Datum und Uhrzeit der letzten \"Anderung \\ {\scshape privs} &amp; Zeichenkette die K\"urzel f\"ur die Benutzerrechte auf dieses Objekt enth\"alt \\ {\scshape commenttype} &amp; Typ des Kommentars \\ {\scshape comment} &amp; Kommentar zum Objekt, wenn vorhanden \\ {\scshape id} &amp; Die Nummer des Repository Objektes \\ {\scshape name} &amp; Der Name des Objektes \\ {\scshape creator} &amp; Name des Benutzers der dieses Objekt angelegt hat \\ {\scshape create\_time} &amp; Datum und Uhrzeit der Erstellung \\ \end{longtable}  \end{document}   The problem is that I don't find a method to force a page break earlier. A page that isn't filled entirely, followed by a page with the last part of a table is better than this.  I tried to add "\needspace{someLength}" at the beginning or end of the rows, but that doesn't work either. It simply makes the rows someLength high instead of forcing a page break if there's no more than someLength room left on the page.  I'd be more than grateful for some solution or even a good hint.  TIA What is the idiom for "as anyone can see" or "clearly"? I keep wanting to say "the handwriting on the wall," but that's not correct. The place I'm trying to use it is, "Alice likes Bob, Carol likes Bob, Alice thinks Bob likes her, but Bob dates Carol, and so as anyone can see, Bob did not like Alice."  What is the appropriate idiom here? Ubuntu 20.04 on DualBoot fail 90% of tries - Initramfs unpacking failed: Decoding failed (c) Let $\mathcal{L}^*\subseteq\mathcal{L}^{\text{FOPC}}$ be a first order language of predicate calculus consisting of three unary function symbols $f,g,h$.  (i) Write down an $\mathcal{L}^*$-sentence $\phi$ such if $\mathcal{A}$ is an $\mathcal{L}^*$-structure with $\mathcal{A}\vDash\phi$ then the restriction of $h_\mathcal{A}$ to the range of $f_\mathcal{A}$ gives a one-to-one function onto the range of $g_\mathcal{A}$.  (ii) Write down an $\mathcal{L}^*$-sentence $\psi$ such that if $\mathcal{A}$ is an $\mathcal{L}^*$-structure with $\mathcal{A}\vDash\psi$ and the domain $A$ of $\mathcal{A}$ is finite then the cardinality $|A|$ is a square.    Part (i) is easy enough, but I'm having trouble with part (ii). I assume that you can use part (i). My initial idea was to use $h$ to ensure $|Range(f_A)| =|Range(g_A)|$ but I'm not sure where to go from there. Thanks. It would be easier to ask for a user's password only once during registration.  The problem: The user could make a mistake while typing the password once because of hiding letters.  The solution: The user could have a toggle button for showing or hiding the password.    Working example with . This approach could be used on the registration or login page.   Are there any benefits to asking a user's password twice during registration vs just not masking the password? Why would you ask twice?  P.S. :        Users make more errors when they can't see what they're typing while filling in a form. They therefore feel less confident. This double degradation of the user experience means that people are more likely to give up and never log in to your site at all, leading to lost business. (Or, in the case of intranets, increased support calls.)   The more uncertain users feel about typing passwords, the more likely they are to (a) employ overly simple passwords and/or (b) copy-paste passwords from a file on their computer. Both behaviors lead to a true loss of security.      Update: I created a . So you may use it if you want to.    Update 2: .  Update 3: Internet Explorer 10 added a toggle password visibility icon. It looks like this:    Update 4: Article about .  Update 5: Example with . Cycles in permutations of a given size how to use \input{file.tex} correctly? Suppose I have two latex files file1 and file2 with the same preamble and so on. Then I want to create a file called bigfile which just puts the two files together. This is not working for some reason. Therefore I first tried to include just file1, this is what I have done:  \documentclass .... %% same as file1 ... same preamble as file1 \begin{document} \input{file1} \end{document}   The result is that I not only get file1, but also the preamble and so on. This is my first experience with putting files together in 1 file. Kindly provide me a solution :) selectively make contact groups silent I'm designing a small module that breaks out a DIP SPI EEPROM to a 2x5 pin header socket(more info here: ). I want to mount the socket on the end of the board, with the pins lying parallel to the board.  I think two 5x1 surface-mount header footprints, on both sides of the board right near the end would do the job. Trouble is, KiCAD doesn't come with such a footprint(if it does, I haven't found it yet). Where might I find one? If one doesn't exist, that's OK; I'll have a shot at making my own. How to change boot splash screen in 18.04 I have done this in the past in 16.04 and 17.10 but in 18.04 I can not get plymouth manager to work.  Downloaded it from the software center but it doesn't work.  Is there a guide available that will let me do it in terminal? Proving the Lorentz invariance of the Lorentz invariant phase space element I have been looking around for a satisfactory answer to prove that  $$\frac{d^3\vec{p}}{2E_{\vec{p}}}$$  where $E_{\vec{p}}=+\sqrt{(|\vec{p}|c)^2+(mc^2)^2}$, is Lorentz invariant. The standard answer seems to be that the above measure is equal to  $$d^4p \, \delta (p^2-m^2)\theta(p_0)$$  where $p^2=p_0^2-|\vec{p}|^2$, and $\theta(x)$ is the step function. I understand these two are equivalent, but I don't understand why the second has to be Lorentz invariant, in particular why the Dirac delta has to be Lorentz invariant. I have found a  (section 2.1) that proves that $\delta^{(4)}(p-p')$ is Lorentz invariant, but I can't find a way to extend their method successfully here. In fact, all I can seem to get tells me the above isn't Lorentz invariant, and that in fact it should transform to   $$\frac{d^3\vec{p}'}{2\gamma E_{\vec{p}'}}$$  which makes sense from $d^4p$ being Lorentz invaraint and $dp_0$ transforming proportional to $\gamma$, but it isn't what everyone else says. What's the problem here?    An alternative way to "derive" the factor of $1/\gamma$:  $$\delta (p^2-m^2)\theta(p_0)=\frac{1}{E_{\vec p}}\delta(p_0-E_{\vec{p}})$$  Now, as $dp_0$ transforms to $\gamma dp_0$, the $\delta(p_0-E_{\vec{p}})$ should transform to $\gamma^{-1}\delta(p_0-E_{\vec{p}})$, as can be shown with an analogous argument to the one shown in the . The $E_{\vec{p}}$ also transforms to $\gamma(E_{\vec{p}}-v p_x)$ for an $x$-boost, but this doesn't seem to solve it. Finding a distribution from the sum of two random variables in a joint trinomial distribution Let X and Y have a trinomial distribution with join probability function:  $ f(x,y) = \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}; x=0,1,...n, y=0,1,...n$  Let $T=X+Y$  I can't seem to figure out where to go exactly. Here's where I am at:  $\sum_{x=0}^{n} \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}\mbox{ substitute in Y=T-X}$ $\sum_{x=0}^{n} \frac{n!}{x!(t-x)!(n-t)!}p^x q^{t-x} (1-p-q)^{n-t}$   I think I could I figure out the algebra.. if I could just see what to do with the summation. I know I have to involve a t somewhere, but I'm not sure. Maybe somebody could just show me how to set up my summation correctly with the substitution? Confidence interval interpretation in linear regression when errors are not normally distributed Ran out of inodes - 100% Full I am using Ubuntu 16.04 on AWS EC2 instance which has been working fine until this problem.  My problem is that I've used 100% of available inodes, as I discovered with this command:  $ df -i Filesystem     Inodes  IUsed  IFree IUse% Mounted on udev           124464    360 124104    1% /dev tmpfs          126787    449 126338    1% /run /dev/xvda1     524288 522638   1650  100% / tmpfs          126787      1 126786    1% /dev/shm tmpfs          126787      5 126782    1% /run/lock tmpfs          126787     16 126771    1% /sys/fs/cgroup tmpfs          126793      4 126789    1% /run/user/1000   The main culprits I discovered:  $ find / -xdev -printf '%h\n' | sort | uniq -c | sort -k 1 -n    1155 /usr/src/linux-headers-4.4.0-62/include/linux    1155 /usr/src/linux-headers-4.4.0-65/include/linux    1155 /usr/src/linux-headers-4.4.0-66/include/linux    1155 /usr/src/linux-headers-4.4.0-70/include/linux    1155 /usr/src/linux-headers-4.4.0-71/include/linux    1155 /usr/src/linux-headers-4.4.0-72/include/linux    1155 /usr/src/linux-headers-4.4.0-75/include/linux    1155 /usr/src/linux-headers-4.4.0-78/include/linux    1155 /usr/src/linux-headers-4.4.0-79/include/linux    1155 /usr/src/linux-headers-4.4.0-81/include/linux    1155 /usr/src/linux-headers-4.4.0-83/include/linux    1155 /usr/src/linux-headers-4.4.0-87/include/linux    1155 /usr/src/linux-headers-4.4.0-89/include/linux    1155 /usr/src/linux-headers-4.4.0-91/include/linux    1155 /usr/src/linux-headers-4.4.0-92/include/linux    1514 /usr/src/linux-headers-4.4.0-62-generic/include/config    1516 /usr/src/linux-headers-4.4.0-65-generic/include/config    1516 /usr/src/linux-headers-4.4.0-66-generic/include/config    1516 /usr/src/linux-headers-4.4.0-70-generic/include/config    1516 /usr/src/linux-headers-4.4.0-71-generic/include/config    1516 /usr/src/linux-headers-4.4.0-72-generic/include/config    1516 /usr/src/linux-headers-4.4.0-75-generic/include/config    1517 /usr/src/linux-headers-4.4.0-78-generic/include/config    1517 /usr/src/linux-headers-4.4.0-79-generic/include/config    1517 /usr/src/linux-headers-4.4.0-81-generic/include/config    1517 /usr/src/linux-headers-4.4.0-83-generic/include/config    1517 /usr/src/linux-headers-4.4.0-87-generic/include/config    1517 /usr/src/linux-headers-4.4.0-89-generic/include/config    1517 /usr/src/linux-headers-4.4.0-91-generic/include/config    1517 /usr/src/linux-headers-4.4.0-92-generic/include/config    1517 /usr/src/linux-headers-4.4.0-93-generic/include/config   Is it safe to delete these files, and if yes could you please tell me the exact command or commands to use on my PuTTY client?