Why do I get an error with this command in KDE:  qdbus org.kde.klipper /klipper getClipboardContents qdbus: could not exec '/usr/lib/x86_64-linux-gnu/qt5/bin/qdbus': No such file or directory   ?
When I run followind command: qdbus I get following error:  qdbus: could not find a Qt installation of ''  I've read alot of old questions with same error but not one helps me.  I've tried execute following commands sudo apt-get install qt5-default + same for for qt4, etc. but still have same error.  If any additional information required for this issue - please add comment and I'll provide info.
Example where union of increasing sigma algebras is not a sigma algebra How do I get a tourist visa/card for Cuba? How do I get a tourist visa (or a tourist card) for Cuba?  I live in the Netherlands, but I'm not a Dutch citizen, I have an Italian passport. What would I do if I want to have a generic method that only accepts types that have overloaded an operator, for instance the subtraction operator. I tried using an interface as a constraint but interfaces can't have operator overloading.   What is the best way to achieve this? How do you disable browser autocomplete on web form field / input tags? How do you disable autocomplete in the major browsers for a specific input (or form field)? Is it really possible to break the speed of light by flicking your wrist with a laser pointer? Hibernating Ubuntu on EC2 instance? I've got an application I'd like to move to EC2. It will likely spend more than half the day completely dormant, so I'm trying to come up with a good solution for starting and stopping it as needed. It takes a few minutes to start up from nothing, so it would be nice if i could hibernate the OS for quicker resumes. I've seen a couple forum discussions on the topic of hibernation within EC2, but never anything conclusive. Has anyone found a working solution to this, or at least some resources that could help me out? Differences between no child vs no children a) They have no child. b) They have no children.  Are both a) and b) correct? Though b) seems more common. There have been  for a while, but this is ridiculous. It reminds me of , but this time it's network-wide.   The banner must be scrolled past (or dismissed) on every page unless both first-party JavaScript and first-party cookies are enabled. If first-party JavaScript is enabled, and the banner is visible, the rest of the page jumps up and down every so often.  Under certain circumstances (not quite sure what they are, yet – a direct link to an answer, maybe? I'm seeing it ) the entire page jumps up and down a line every few seconds even when the banner is off-screen.   The cookie banner keeps coming back unless both third-party JavaScript  are enabled (and  is solved). The cookie box occludes the “Learn more about Teams” link in the banner (with default Firefox window sizes) and the Job / Teams link in the sidebar.  As it stands,  – a site for hobbyists and researchers, whose audience is not very likely to purchase Teams – is much, much harder to use, for no benefit that I can see. Can you roll back this change, please? Is it two-third or two-thirds? Is it two-third or two-thirds? If both are correct, how does their usage differ? Best way to center a  on a page vertically and horizontally? My Mac is getting really slow, what should I do? How to view the moderators here in MSE? I  am able to view the Stack Overflow site moderators by clicking on this    But when I search the same in meta.stackexchange, it does not display the site moderators.    The below message is displayed      We believe moderation starts with the community itself, so in addition   to privileges earned through reputation, we periodically hold   democratic moderator elections.      Please see A Theory of Moderation for information on our moderation   philosophy   My Questions:   Are there restrictions to view the site moderators in meta.Stackexchange? How does the above message make sense when viewing the moderators? Can someone explain this to me? What's wrong with this equal probability solution for Monty Hall Problem? I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.   Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.   Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)? Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.   Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.   What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks. I want to write as a math fraction 1/4 " ( a quarter of inch). I want to write this fraction out of an equation because is a characteristic of an element. When has Wonder Woman died? I know Wonder Woman has died in canon before, but in which issue did this happen? Prime dividing the binomial coefficients It is quite easy to show that for every prime $p$ and $0&lt;i&lt;p$ we have that $p$ divides the binomial coefficient $\large p\choose i$; one simply notes that in $\large \frac{p!}{i!(p-i)!}$ the numerator is divisible by $p$ whereas the denominator is not (since it is a product of numbers smaller than $p$ and $p$ is prime).  My problem is with generalizing this argument for $q=p^n$. I'm looking for the most elegant and simple way to prove that $p$ still divides $\large q\choose i$. Join matrix based on conditions Relationship between Partial Harmonic Sum and Logarithm. How do I pull from a Git repository through an HTTP proxy? Note: while the use-case described is about using submodules within a project, the same applies to a normal git clone of a repository over HTTP.  I have a project under Git control.  I'd like to add a submodule:  git submodule add http://github.com/jscruggs/metric_fu.git vendor/plugins/metric_fu   But I get  ... got 1b0313f016d98e556396c91d08127c59722762d0 got 4c42d44a9221209293e5f3eb7e662a1571b09421 got b0d6414e3ca5c2fb4b95b7712c7edbf7d2becac7 error: Unable to find abc07fcf79aebed56497e3894c6c3c06046f913a under http://github.com/jscruggs/metri... Cannot obtain needed commit abc07fcf79aebed56497e3894c6c3c06046f913a while processing commit ee576543b3a0820cc966cc10cc41e6ffb3415658. fatal: Fetch failed. Clone of 'http://github.com/jscruggs/metric_fu.git' into submodule path 'vendor/plugins/metric_fu'   I have my HTTP_PROXY set up:  c:\project&gt; echo %HTTP_PROXY% http://proxy.mycompany:80   I even have a global Git setting for the http proxy:  c:\project&gt; git config --get http.proxy http://proxy.mycompany:80   Has anybody gotten HTTP fetches to consistently work through a proxy?  What's really strange is that a few project on GitHub work fine ( for example), but others consistently fail ( for example). Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?  Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.  If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.