64 bit. Title explains it all. Just want Firefox beta.
I recently discovered that my computer is 64-bit. I have backed up my Home folder, and reinstalled Ubuntu. The reinstall wasn't nearly as painful as I thought. There is one thing that I can't quite seem to figure out: how do I get the 64-bit Firefox Beta build? I always get the Beta builds, but I want to take advantage of the 64-bit architecture of my computer.  says that Mozilla has come out with a 64-bit version of Firefox, but I can't seem to find it.  I do understand  of using a 64-bit browser, but I've decided to jump right in and do it anyway. (Flash and Java are already 64-bit, and who cares about Silverlight, since it's not for Linux anyway?)  There's only one issue, and it's a big one: I can't find the 64-bit Beta PPA!!! (I really hate using .tar.gz files, but I'd be willing to do that as long as I could still access Firefox via the Launcher. Oh, speaking of which, I don't understand .tar.gz files. Once, I managed to run one (the Dropbox Beta build,) but I have no idea whatsoever on how to install them: as in, click on the icon and go.)
Solving SDE: $dX(t) = udt + \sigma X(t)dB(t)$ Why didn't Gollum grow old and die after he lost the One Ring to Bilbo? How can I use wget to get files recursively if the username contains an @? Firstly my main goal is to try and get FTP files recursively in Centos.  However the actual FTP login contains an @ which ruins the syntax.  Is there some way to place the username in "double quotes" or some such equivalent such that I can use @in the username? terminator: permission denied for sudo Is there a way to indicate to  how/where to find (paths to)  tools such as python, nmap, php and so on?  I ask this because when I schedule a job using crontab -e, I always need to specify the full path of python, nmap and whatever other tool I want to use. This is from page 3 of  ().  Copying the relevant segment:     Stirling’s approximation tells us $\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n$. In particular we can use this to say that   $$ {n \choose k} \leq \left(\frac{en}{ k}\right)^k$$   I tried the tactic of combining bounds from $n!$, $k!$ and $(n-k)!$ and it didn't work. How does this bound follow from stirling's approximation? Pgfplots: How to fill area under a curve in a 3d plot similar to \closedcycle in 2d Let $R$ be an rng. (There may be no unity)  Then, does there always exist a ring(with unity) $A$ such that $R$ is a subrng of $A$? Proving $T$ and $U$ are equal if I know $(X,T)$ and $(X,U)$ are hausdorff and compact This question is from Wayne Patty Exercise 4.2  Let $T$ and $U$ are Topologies on set X. Suppose $ (X,T) $ and $(X,U)$ are compact Haurdsorff Spaces. Prove that $T=U$ or $T$ is not a subset of $U$ or $U$ is not a subset of T.  I thought of proving it by this way: Assuming that if $T$ is a subset of $U$ or $U$ is a subset of $ T$ then $T=U$ must be true. But I was unable to think on how to use compactness or hausdorff property and was unable to proceed. Well, also this question was asked before on this website. But I am unable to understand how the solution will work despite trying to understand it for a fair amount of time as it was just outlined. So, I am looking for a detailed solution. Node Package Manager got corrupted in some way, now it cannot be installed Can I clear an area of my prison for everyone BUT max security prisoners? How can I make sure that only low and medium security inmates are in a specific area? Is it possible to reverse the overlay you can set? It seems like you are only able to clear areas for low OR medium security prisoners, but sharing is not possible. The thing I want here is to make sure that max security prisoners are not in certain areas. Let $(X, Y)$ be distributed as a multivariate normal with parameters  $$ \mu = \begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix} \qquad \Sigma = \begin{bmatrix} \sigma_X^2 &amp; \sigma_{XY} \\ \sigma_{XY} &amp; \sigma_Y^2 \end{bmatrix}. $$   I would like to calculate $E(X | y_1 &lt; Y &lt; y_2)$, where $y_1$ and $y_2$ are constants.  From , I have managed to work out that $$ E(X | Y &gt; y_1) = \mu_X + \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}{1 - \Phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}\right]\\ E(X | Y &lt; y_2) = \mu_X - \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}{\Phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}\right] ,$$ where $\phi(\cdot)$ and $\Phi(\cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 &lt; Y &lt; y_2)$, though.  I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them: Please help me to answer the following problem:  Let $f(x)=x^3-3x-5\in\mathbb{Q}[x]$  Thanks Divisibility by 7 What is the fastest known way for testing divisibility by 7? Of course I can write the decimal expansion of a number and calculate it modulo 7, but that doesn't give a nice pattern to memorize because 3 is a primitive root. I'm looking for alternative ways that can help you decide when a number is divisible by 7 by hand.  I'm sorry if this is a duplicate question, but I didn't find anything similar on the site. How would I solve the following problem involving elementary matrices? Let $A=\begin{pmatrix}0&amp;5\\ 7&amp;4\end{pmatrix}$.      1) Write $A$ as a product of $4$ elementary matrices.      2) Write $A^{-1}$ as a product of $4$ elementary matrices.   My work. I have managed to find $A^{-1}$, which came out to be this:  \begin{pmatrix}-\frac{4}{35}&amp;\frac{1}{7}\\ \frac{1}{5}&amp;0\end{pmatrix}  However, I am struggling to figure out how I would split each of these matrices into $4$ elementary matrices. Any help? In cricket, in UDRS screen, if the ball hits the wicket, then why is not out still given? In cricket, in UDRS screen, if the ball hits the wicket, then why is not out still given?  I saw this situation so many times happened; bowler appealed for UDRS for LBW, in UDRS it shows ball hits the wicket, but still not out is given sometimes. Access to journals for individuals outside of an institution? For all that is said about the wealth of information that is freely available on the internet, the fact is that most of it is incomplete, dumbed down, lacking in context or downright wrong.   Individuals spend the best part of 20 years completely oblivious to the existence of journals and the process of peer review and even then we'd be lucky if 10% of those with access to journals (usually through an institution) actually bother to invest in them.   Granted that some journals will be totally inaccessible to those without a solid background in some particular field but by and large exposure is a good thing and there are many areas of research where individuals will benefit directly from having read these.  Where does this come from?  I decided I wanted to read Popular Politics in the Late Medieval City: York and Bruge in The English Historical Review today, mainly because it looks interesting and I'm hoping to invest more time into learning about British and European History. Incidentally I also wouldn't mind tucking into Super Stable Clocks, Nature 500, 505 (29 August 2013) and a number of data analysis and big data journals in order to advance my career.   Now, not being affiliated with an institution that subscribes to these journals means that I would typically have to pay between £5 ($10) and £15 ($25) per article that I read... I could probably spend £100 just to pass an hour by.  Question   What's being done to address the lockdown on 'higher' education material? I've noticed that the Directory of Open Access Journals publishes open articles and other firms are doing similar, mainly with journals in new fields? How could I, as an individual outside of an institution, gain journal access without it costing so much money? A graph $G$ is bipartite if and only if $G$ can be coloured with 2 colours How does PK know better than Jaggu? In , PK knows about Jaggu's boyfriend that he didn't cheat/left her in Belgium. But PK only read the mind of Jaggu. PK didn't know about Tapasvi's lies. How does he know that Sarfaraz still loves Jaggu in the ending? GNOME: supend on laptop lid close no longer works since 19.10 upgrade I'm quite positive that this used to work flawlessly in 19.04 (and past releases) but no longer does after a release-upgrade to 19.10 (kernel 5.3.0-24, ThinkPad X1 Carbon gen 5 in case this matters).  I have the corresponding switch set in GNOME-Tweaks and have also attempted to manually edit /etc/systemd/logind.conf as suggested , namely uncomment and adjust the following lines, followed by a reboot:  $ grep -v '^#' /etc/systemd/logind.conf [Login] HandleLidSwitch=suspend HandleLidSwitchExternalPower=suspend HandleLidSwitchDocked=suspend LidSwitchIgnoreInhibited=yes   (strictly speaking, only the first one ought to be required but, just in case, I also added lines 2 &amp; 3, as well as #4 for good measure...  Looking at the output of systemctl status systemd-logind, I see the following:  Dec 17 17:56:50 x1c5 systemd-logind[8959]: Lid closed. Dec 17 17:57:03 x1c5 systemd-logind[8959]: Lid opened.   So the lid events are correctly detected.  To rule out any issue with the suspend process itself, I have also set my power button to suspend in Settings->Power->Suspend &amp; Power Button, for testing purposes.  Pressing the power button now results in this in the output of systemctl status systemd-logind (after a successful suspend-resume cycle):  Dec 17 18:13:58 x1c5 systemd-logind[8959]: Power key pressed. Dec 17 18:14:28 x1c5 systemd-logind[8959]: Delay lock is active (UID 121/gdm, PID 9215/gsd-power) but inhibitor timeout is reached. Dec 17 18:14:39 x1c5 systemd-logind[8959]: Operation 'sleep' finished.   Bottom line: lid events are detected correctly and suspend works flawlessly - but not on lid close...  Any idea how I can get the suspend on lid-close feature back to life?  I can of course use the workaround with the power button but it's a little too easy to forget and result in laptop overheating in bag etc...  --- edited to add ---  Got a little further courtesy of , although the issue isn't the exactly as described.  Here's a smoking gun:  $ systemd-inhibit --list --mode=block WHO       UID  USER      PID  COMM           WHAT                                                     WHY                                                         MODE  gdm       121  gdm       2231 gsd-power      handle-lid-switch                                        External monitor attached or configuration changed recently block gdm       121  gdm       2252 gsd-media-keys handle-power-key:handle-suspend-key:handle-hibernate-key GNOME handling keypresses                                   block user      1000 user      3017 gsd-media-keys handle-power-key:handle-suspend-key:handle-hibernate-key GNOME handling keypresses                                   block   This suggests that the lid close action is being blocked because "External monitor attached or configuration changed recently".  I do connect my laptop to a monitor using a USB-C cable but as I write this, it's been undocked and running on batteries for nearly 3 hours (GNOME settings only shows my laptop display as available).  However it looks like systemd or whichever component is responsible for this still incorrectly believes the external display is attached and - per the ticket - suspend isn't supported by GNOME in such scenarios...  Interestingly the gsd-power process (PID 2231) is owned by gdm rather than my UID.  Killing it restores the suspend-on-lid-close functionality.  Need to figure out why this process exists... When actually connected to external display, systemd-inhibit --list --mode=block reports a 2nd gsd-power process owned by the interactive user; which appears correct and sufficient.