I am using Ubuntu 14.04. I am unable to hibernate my system with sudo pm-hibernate. The screen only flashes then it returns to desktop, nothing happens. I have a swap partition of 4GB my RAM size is 3GB. Even the hibernate tray option doesn't actually hibernate the system instead only sends the system to sleep. Kindly help.
I'm using Xubuntu 14.04 64 bit. I know that question been asked a billion times, but all of them consist of editing following file:  /etc/polkit-1/localauthority/50-local.d/com.ubuntu.enable-hibernate.pkla   That file does not exist on my system. I tried com.xubuntu.enable-hibernate.pkla but doesn't exist either, locate command doesn't find it.
Difference between ' and " on command line (bash)? Strictly diagonally dominant matrices are non singular I'm trying to understand an algorithm that tries to map points from a lie group to its lie algebra using the exponential map. The background is the representation of 3d coordinate transformations as a lie group.  I can imagine, how one can map points on a manifold (eg a sphere) to the tangent space (tangent plane) at some point x on the sphere. I can imagine to create the tangent space by computing derivatives at that point x. But I cannot imagine how the exponential of some matrix is giving me that tangent space?  How can I envision the exponential of a matrix? Is there some picture or analogy I can use as intuition? I was so excited about the Spring '14  as a way to present data visualizations on a public site, and now it looks like that's not possible.  I created a page that looks basically like the release notes sample:  &lt;apex:page &gt;     &lt;analytics:reportChart reportId="00OW0000000Spru"&gt;&lt;/analytics:reportChart&gt; &lt;/apex:page&gt;   I tested it inside the authenticated environment, where it works fine, then added it to a Site. There, I get the error: To view the report chart, you need both the Run Reports permission and access to the report's folder. But, Run Reports is not a permission I can grant to the Site Guest User.  Anybody have any ideas for a workaround? Or is it not going to be possible to show report charts on a public site? Lie Algebra of $SL_n(\mathbb H)$ The Lie algebra of $SL_n(\mathbb C)$ are the matrices where the trace is $0$. But what is the Lie algebra of $SL_n(\mathbb H)$ where $\mathbb H$ is the quaternions? Why crontab scripts are not working? A PC in our group was off on his own following a recurring villain of our groups. He was paralyzed where he could move, and the vampire bit his neck. The vampire said "now you get to remain 19 (years old) forever" before he left the cave. This was all done in RP.  We found him later, and healed up his bite marks. Our characters know of vampires, specifically only the one that bit him (recurring villain), but don't really know how vampirism works, because it isn't a super common thing in this world yet.   My question as the player is:  How does RAW vampirism work in 5e?  Is it a disease that develops over time? Is it something where they have to die and be brought back as a vampire? $k$ times's draw from $n$ numbers with replacement, each number at least appear once Sample $k$ times with order from $n$ distinct numbers with replacement where $k\ge n$. Here "with order" means the sample is treated like a sequence rather than a set, so results like "1,3,3" and "3,1,3" are different.  The question is what is the probability that every number is sampled at least once? My answer is $1-\frac{{\sum\limits_{i = 1}^n {C_n^i\times i! \times {{(n - i)}^k}} }}{{{n^k}}}$ where $C_n^i$ denotes $n$ chooses $i$.  ${C_n^i\times i! \times {{(n - i)}^k}}$ means first choose $i$ ($i=1,2,...,n-1$) of numbers not to appear in the sample, and then we sample from the remaining $n-i$ numbers for $k$ times.   My question is, is there more elegant solution where the numerator can be directly computed? i.e. can we directly calculate the number of possible sample sequences in which every number has at least one occurrence? Thank you!    PS: my attempt was $C_k^n\times n!\times n^{n-k}$, meaning that we first choose $n$ positions in the sequence of length $k$ to place $n$ numbers, and then permute the $n$ numbers. This is to ensure every number to appear once. The remaining $n-k$ are sampled from the $n$ numbers without restriction.   However, I later found this seems not correct, and it counts the same sequence multiple times.  PS2: as pointed out in the comments, my first answer "1- ..." is not correct either. Determining the number of bit strings of length n that contains no adjacent 0s • $C_n =$ this number of bit strings  • A binary string with no adjacent 0s is constructed by   Adding “1” to any string w of length $n-1$ satisfying the condition, or Adding “10” to any string v of length $n-2$ satisfying the condition   I still don't find that intuitive at all. Why adding up $C_{n-1} + C_{n-2}$ results in $C_n$? I have tried my best and asked my friends for help, but they find the formula unintuitive as well. Could anyone here come up with a clear explanation? Thanks! In comments, moderators have their diamonds in the OP rectangle:        And in the review (suggested edits), mods have their diamonds outside the blue OP background highlight rectangle:        Of course, it Doesn't Really Matter At All, but it is a small discrepancy (I am not saying that in the review, OP rectangles have a bigger padding). What follows if we fail to reject the null hypothesis? What conclusions can we draw if $p&gt;\alpha$? Does not rejecting the $H_0$ mean anything? pgfplots -> contour gnuplot with lualatex instead of pdflatex Maximum Likelihood Estimate with different parameters Suppose that X and Y are independent Poisson distributed values with means $\theta$ and $2\theta$, respectively. Consider the combined estimator of $\theta$ $$ \tilde{\theta} = k_1 X + k_2 Y $$ where $k_1$ and $k_2$ are arbitrary constants.   Find the condition on $k_1$ and $k_2$ such that $\tilde{\theta}$ is an unbiased estimator of $\theta$. For $\tilde{\theta}$ unbiased, show that the variance of the estimator is minimized by taking $k_1 = 1/3$ and $k_2 = 1/3$. Given observations $x$ and $y$ find the maximum likelihood estimate of $\theta$ and hence show that $\tilde{\theta}$ is also the maximum likelihood estimator.   I have gotten (1) and (2) okay, but it's (3) I am having trouble with, I'd be okay if $X$ and $Y$ had the same parameter but I'm having trouble with $X$ and $Y$ having different parameters, any help would be appreciated.   NOTE   For (1) I got $k_1 = 1 - 2k_2$.   For (2) I found the variance of $\tilde{\theta}$, then differentiated and let equal to zero to minimize - therefore we get (after subbing in $k_2 = 1 - k_1/2 $) $$3k_1-1=0,$$ which when subbing in $1/3$, we see it is minimised.   Thank you. Can anyone can suggest where I can freely download ArcObjects SDKs for ArcGIS?  I am using version 10.2. Let $R$ be a commutative ring with identity, and $S$ a multiplicative subset of $R$ not containing $0$. Which step in my reasoning is wrong?   Every element of $S$ is a unit in $S^{-1}R$. Every element of $S$ that's a zero divisor in $R$ is a zero divisor in $S^{-1}R$. But zero divisors can't be units. Therefore it's not possible to localize $R$ at $S$. If you are root, and you issue   rm -rf /    Then how far can the command go? Can you recover data from this kind of an action? Even after the binaries are gone, would the running processes still be active? What would it take to make the same physical machine boot again? What files would you need to restore to make this happen?  I could try this on a VM and see, but I want to know the rationale behind what to expect if I do this. How do we see? Where do the photons disappear? I know that the light is reflected from a object to my eyes, but I don't understand exactly how. The photons appear from the light source and disappear in my eye! Can someone explain the phenomenon of where the photons go and do to allow us to see? NSdate Assuming wrong time zone Why does evolution not make life longer for humans or any other species?  Wouldn't evolution favour a long life? Where should I post my sum-up of the answers, as the OP?