When using multiple monitors in Kubuntu, can I make it more difficult for the mouse to cross screen boundries? When running multiple monitors on Windows and macOS there is a feature that allows you to set a sort of pressure or dead area on screen boundaries between monitors. You have to push through this area with your mouse cursor to get to leave the current screen/monitor and get to the next one.  This makes it easier to stay within the bounds of whatever monitor your mouse cursor is on.  I know KDE/Kubuntu has a Screen Edges setting, but setting/changing the "Toggle Window Switching" or Switch desktop on Edge" options didn't seem to have any effect.    Edit:  I am running Ubuntu 18.04 with KDE Plasma 5.12.6  Output of inxi -Sxxx:  System:    Host: ubuntu-portable Kernel: 4.15.0-23-generic x86_64 bits: 64 gcc: 7.3.0  Desktop: KDE Plasma 5.12.6 (Qt 5.9.5) dm: sddm,sddm Distro: Ubuntu 18.04 LTS  Regarding being marked as duplicate:  This is not a duplicate of the question provided in the link. I understand that there is confusion over the term "sticky edge". The linked question does not address how to enable the feature whereby the mouse cursor clings to the edge in a multi-monitor setup (whatever it is called).   I can find no option in KDE for sticky edges. As posted in my photo, the Active Screen Corners and Edges setting panel seems like the logical place to put this, but I don't see it. Perhaps that is my answer, and that this feature isn't available in KDE.
What is the function of the "Sticky edges" on/off setting in displays? I have installed Ubuntu 12.04 (Precise Pangolin). If I open System Settings &gt;&gt; Displays there is a new (?) on/off button for a setting named Sticky edges.  Can someone tell me what this setting is supposed to control?
How to put a full-width table at the top or bottom of the same two-column page as the reference text? Unexpected workspace change? My friend who has only recently downloaded blender has come across a workspace issue. He said he accidentally pressed a button and it has entirely changed his user interface/ Workspace. He can't open, save or do anything with the change.    This is a full window image! Consider the following sentences:        I'll move to the south by then probably.   I'll probably move to the south by then.   I'll move to the south probably by then.      Which of the three is correct. In my opinion, the third one is wrong. First two are correct with a small difference in their meaning.  First one makes my probability refer to the time when I'll be moving and the second one tells that there is still a chance that I might not move.  Am I correct? In the Harry Potter series, are wizards capable of creating their own spells, or are they limited to the spells that they're taught/learn?  I seem to remember something about Snape creating sectumsempra, but I'm not sure if that really is the case. Draw Vertices of Regular Polygon I am trying to draw just the nodes (no sides) at the vertices of a regular hexagon.  should be easily modified to accomplish what I want, but I am such a novice that I have difficulty understanding the example. Ideally, each vertex would be a separate named node so that I can easily draw edges between them (I will be making several different graphs using these same nodes). Is there any (even the most subtle) difference between the meaning (or connotation) of that is and i.e.?  I used to be convinced that they were synonyms, until an editor of a scientific journal changed all occurrences of i.e. into that is. Probability Interval for Gamma Distribution There is something that I am doing wrong in the exercise below and I would appreciate some help figuring it out.  Let $X_1, X_2, \ldots, X_5$ be a random sample from a $\Gamma \left(3,3 \right)$ distribution . Define $W=\sum_{i=1}^5 X_i $.    Obtain the distribution of $W$ and of $2W/3$ Find $c_1$ and $c_2$ such that $P \left( c_1 &lt; W &lt;c_2 \right)=0.95 $     Since $W$ is the sum of iid gamma variables, $W \sim \Gamma \left(15,3 \right)$ and now a straightforward transformation argument shows that $Y=\frac{2}{3} W \sim \chi^2 \left(30 \right) $. This completes question 1.  Now for question 2, my thoughts were to transform these probabilities using the result from question 1, namely that $\frac{2}{3} W \sim \chi^2 \left(30 \right) $. Hence, unless I have made a horrible mistake here:  $$P \left( \frac{2}{3} c_1 &lt; Y &lt; \frac{2}{3} c_2 \right)=0.95 $$  We can find the constants from a chisquare table for $30$ degrees of freedom. In my case, they are $16.791$ for the lower bound and $46.979$ for the upper one.   Then if we equate $\frac{2}{3}c_1=16.791$ and solve for $c_1$ we should be able to find the constant for the above gamma confidence interval, right?   The problem is that according to $R$, $P \left( W&lt;c_1 \right) =1 $, instead of  the desired $0.025$ which means I have gone wrong somewhere. I do not think it is in question though $1$ as I have verified the result many times.   Thank you. I have a Queue&lt;T&gt; object that I have initialised to a capacity of 2, but obviously that is just the capacity and it keeps expanding as I add items.  Is there already an object that automatically dequeues an item when the limit is reached, or is the best solution to create my own inherited class? How to control fan speed? Showing $\left(1+\frac{1}{n}\right)^n$ is an increasing sequence by comparing the binomial expansions term by term So I know that showing $\left(1+\frac{1}{n}\right)^n$ is an increasing sequence has probably appeared on this site about 100 times, but my professor said he thinks the induction step is most easily seen if you expand this out and compare the summation term by term. I tried this and I did not what see why the inequality was obvious. Can somebody help me out here? Thanks! What do they do with all the plates and cups in Star Trek? Why do we say that irreducible representation of Poincare group represents the one-particle state? Only because   Rep is unitary, so saves positive-definite norm (for possibility density), Casimir operators of the group have eigenvalues $m^{2}$ and $m^2s(s + 1)$, so characterizes mass and spin, and It is the representation of the global group of relativistic symmetry,    yes? Prediction interval for binomial random variable What is the formula (approximate or exact) for a prediction interval for a Binomial random variable?  Assume $Y \sim \mathsf{Binom}(n, p)$, and we observe $y$ (drawn from $Y$). The $n$ is known.  Our goal is to obtain a 95% prediction interval for a new draw from $Y$.   The point estimate is $n\hat{p}$, where $\hat{p}=\frac{y}{n}$. A confidence interval for $\hat{p}$ is straightforward, but I cannot find a formula for a prediction interval for $Y$. If we knew $p$ (rather than $\hat{p}$), then a 95% prediction interval just involves finding the quantiles of a binomial. Is there something obvious I'm overlooking? Problem Statement  Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$   Suppose that the following conditions hold:   $f(1)=0$ the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$      $$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$   My attempt  Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$  Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$  $$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$  And so we have $$a = -(p+q+r) = -p^2q^2r^2$$ $$b = pq + pr + qr = p^2 + q^2 + r^2$$ $$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$  Then we get the following relations  $$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$  From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$  However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?  Edit  For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$ I'm creating a function where I need to pass an object so that it can be modified by the function. What is the difference between:  public void myFunction(ref MyClass someClass)   and  public void myFunction(out MyClass someClass)   Which should I use and why? Two diagrams side by side I am trying to put two feynman diagrams side by side, but they get displayed below eachother. I am using feynmf and subfloat. This normally works for pictures.  \begin{figure}[ht!]       \centering     \subfloat{          \begin{fmfgraph*}(40, 25)         \fmfstraight          \fmfbottom{wm,i2,z2}           \fmftop{wp,i1,z1}           \fmf{dashes}{wp,i1,z1}           \fmf{dashes}{z2,i2,wm}           \fmf{dashes,caption=$W^{+}$}{i1,i2}         %  \fmf{dashes,caption=$Z^0$}{i2,z1}          % \fmf{dashes,caption=$Z^0$}{i2,z2}         \fmfv{lab=$Z^0$}{z1}         \fmfv{lab=$Z^0$}{z2}     \fmfv{lab=$W^+$}{wp}     \fmfv{lab=$W^-$}{wm}         \end{fmfgraph*}     }      \subfloat{          \begin{fmfgraph*}(40, 25)         \fmfstraight          \fmfbottom{wm,i2,z2}           \fmftop{wp,i1,z1}           \fmf{dashes}{wp,i1,z1}           \fmf{dashes}{z2,i2,wm}           \fmf{dashes,caption=$W^{+}$}{i1,i2}         %  \fmf{dashes,caption=$Z^0$}{i2,z1}          % \fmf{dashes,caption=$Z^0$}{i2,z2}         \fmfv{lab=$Z^0$}{z1}         \fmfv{lab=$Z^0$}{z2}     \fmfv{lab=$W^+$}{wp}     \fmfv{lab=$W^-$}{wm}         \end{fmfgraph*}     }      \caption{Exchange of W boson}     \end{figure} How to suppress Terminated message after killing in bash? Should the first word after a colon be capitalized? Should the first word after a colon be capitalized? Which of the following is correct?   For example: This. For example: this. I want to animate a character in a realistic but easy way. I was thinking that I could use videos of me performing the actions I want to have on my characters, however I can't find a way of placing the videos and make the character to follow my movements in an automated way, like motion tracking. How can I animate automatically using this technique? Are there any Plug-ins to work like this? Is this mehod actually more useful than standard animation techniques? What features of Mavericks are beneficial to you? This question serves to share and collect the enhancements that make a large difference to how you use your Mac.   Please post one feature per answer. Please also check to see if your answer has already been posted - duplicate answers will be deleted. To search answers for this question use  (directly from the question page) in addition to your search terms in the search box in the upper right hand corner of this page.  The best answers will not only list a feature, but provide details on how to configure that feature, and provide an image of how to use the feature to be more efficient or effective with Mavericks.