I have installed VMware on Ubuntu 13.10 but when i launch it then a message said that i  have to specify the path for the linux-headers-3.11.0-19-generic. Now i'm not expert on linux, i give the command "uname -a" here the result:  Linux MarioP 3.11.0-19-generic #33-Ubuntu   My knowledge of linux kernel stops here. Can you help me?
I installed the VMware bundle on my Ubuntu 11.04 successfully but when I open it it gives me this window     and I don't know the path to this C headers.
Seasonal Loan Amortization Can the Eclipse Java formatter be used stand-alone Is there a way to use the formatter that comes with eclipse, outside of eclipse?  I would like to format some java files using my formatter.xml file that I have configured using eclipse.  Does anyone have any code examples that would allow me to do this?  I would also like to use this standalone, so the specific jars that are used would be nice. Suppress "In:" biblatex How do I suppress the "In: " before the journaltitle? Like so:     In: Journal of Applied Physics How to get the file name from a full path using JavaScript? Is there a way that I can get the last value (based on the '\' symbol) from a full path?  Example:  C:\Documents and Settings\img\recycled log.jpg  With this case, I just want to get recycled log.jpg from the full path in JavaScript. Monotonicity of the sequences $\left(1+\frac1n\right)^n$, $\left(1-\frac1n\right)^n$ and $\left(1+\frac1n\right)^{n+1}$ Why is "cupboard" pronounced with a silent "p"? After offering a bounty to , I realize that I cannot comment on the question because I no longer have the required reputation. Is it possible to change this behavior, so that the bounty owner can comment only on the post with the bounty irrespective of their reputation? While bounty owners are told that they'll lose privileges at the time they place a bounty, it makes sense to allow them to interact with potential answerers/commentators to help solve the problem, or at least provide additional information. How does ServerName and ServerAlias work? If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1 + d(x,y)}$ is also a metric Let $(X,d)$ be a metric space and for $x,y \in X$ define $$d_b(x,y) = \dfrac{d(x,y)}{1 + d(x,y)}$$ a) show that $d_b$ is a metric on $X$  Hint: consider the derivative of $f(t)$ = $\dfrac{t}{1+t}$  b) show that $ d$ and $ d_b $ are equivalent metrics. c) let $(X,d) $ be $(\mathbb{R}, |\cdot|)$ Show that there exists no $ M&gt;0$ such that $ |x-y| $ $\leq$ $Md_b$ $(x,y)$ for all $ x,y$ $\in$ $\mathbb{R}$  I have calculated $f(t)$ and $f'(t)$ and from this I know $f(t)$ is an increasing function as $f'(t)$ is strictly positive. But I don't know where to go from here or how to do parts b) and c) Collinearity can pose certain problems in various kinds of regression problem. In particular, it can make the parameter estimates have high variance and be unstable.   Various methods have been proposed to deal with this including ridge regression, partial least squares regression, principal components regression, dropping variables and getting more data.   One method that is controversial is standardizing or scaling the independent variables, with various experts saying it is a good (e.g. Garcia) or bad (e.g. Belsley) idea. Belsley's issue seems to be (in non-technical terms) that changing the IVs just pushes the problem under the rug.  But other experts seem not to agree. And the authors tend to get rather heated in defending their own positions.   Back when I did my dissertation (on collinearity diagnostics) I found Belsley's arguments convincing, but that was long ago (I got my degree in 1999).  I'm looking either for expert guidance or for any current review article that is unbiased. Explicit "do not award bounty" button Shouldn't there be a way to prevent automatic awarding of a bounty? For example, if someone provides an answer that sounds good but is totally wrong, it may happen that it gets a score of +2, but doesn't deserve the bounty. If no one else answers, half the bounty will automatically go to this person. Automatic awarding is good if an inactive user offers a bounty, but I think there should be a button that basically means "I know the bounty should now be awarded, but I have read the answers and none of them deserve the bounty".  The button should obviously not refund the bounty, just let it go to /dev/null instead of awarding it to someone who doesn't deserve it. Not enough free disk space when upgrading Significant predictors become non-significant in multiple logistic regression When I analyze my variables in two separate (univariate) logistic regression models, I get the following:  Predictor 1:    B= 1.049,    SE=.352,    Exp(B)=2.85,    95% CI=(1.43, 5.69),    p=.003    Constant:    B=-0.434,    SE=.217,    Exp(B)=0.65,                            p=.046  Predictor 2:    B= 1.379,    SE=.386,    Exp(B)=3.97,    95% CI=(1.86, 8.47),    p&lt;.001    Constant:    B=-0.447,    SE=.205,    Exp(B)=0.64,                            p=.029   but when I enter them into a single multiple logistic regression model, I get:  Predictor 1:    B= 0.556,    SE=.406,    Exp(B)=1.74,    95% CI=(0.79, 3.86),    p=.171 Predictor 2:    B= 1.094,    SE=.436,    Exp(B)=2.99,    95% CI=(1.27, 7.02),    p=.012    Constant:    B=-0.574,    SE=.227,    Exp(B)=0.56,                            p=.012   Both predictors are dichotomous (categorical). I have checked for multicollinearity.  I am not sure if I have given enough info, but I cannot understand why predictor 1 has gone from being significant to non-significant and why the odds ratios are so different in the multiple regression model. Can anyone provide a basic explanation of what is going on? Is SO/SE about to be sold (and does that explain recent events)? Commutator of two normal subgroups with trivial intersections So I tried looking everywhere but I'm really lost with this proof. I need to to show that:  Let $H$, $K$ be normal subgroups of a group $G$, satisfying $H\cap K = \{e\}$. Let $h\in H$, $k \in K$. (a) Prove that the commutator $w = hkh^{-1}k^{−1}$ satisfies $w \in H$ and $w \in K$. Then conclude that $w = e$ (b) Prove that $hk = kh$, for all $h \in H$, $k \in K$. How to backup all my data from a server? How fast would someone have to run to run over water? I was thinking about Flash, the superhero, or the little boy in the Incredibles.   There is  that doesn't answer a lot. Especially, I don't think surface tension would help a lot for a human to run over water, I think one would have to build on the inertial effect of the water.  There is one empirical approach based on figuring out the speed at which bare-footed water skiing is done, but I wasn't able to find a decent number. Still the difference may be that a hypothetical runner would have to propulse herself over the water, which may or may not make the thing more difficult.  Hence, I raise the question here. Stop iPhone alerting me when I'm in a phone call When I'm in a phone call, I still get alerts for emails and texts. The noise will play and, if the phone is on Speaker, it will vibrate -- often leading the person I'm talking to wonder what that weird noise was.  Surely there must be a way to silence the phone while you're talking on it. It's very distracting to have beepings going off right next to your ear! Use Yaml for MongoMapper Config Don't revert reputation from suggested edits when reviewer's account is deleted I noticed something strange in the reputation of one of my accounts today:        Never saw "reverted" event before, so went to  and realized what happened: the user who approved my suggested edit has been removed/deleted their account.  The edit is still there. The post is still there. The effort spent in editing is still there, untouched. I really can't see any reason to take the reputation away in such a case.  Can this please be changed so that the reputation won't be lost? Personally I don't care that much, and got lots of reputation from other sources, but many people have low reputation and each point matters to them.  P.S.  This might be a bug introduced when implementing this recent change about "", so tagging as a bug too.