I modified sudoers. All users can not use sudo now. How should I fix this? I was trying to save times to type password for sudo because one of my scripts had to be randomly repeated quite often.  I googled some information about how to bypass typing password and applied to my server.  Now the sudo account is corrupted and all the users on server cant not use sudo.  How can I fix this?  What I did:   I created mysudoers in /etc/sudoers.d/  sudo vim /etc/sudoers.d/mysudoers Added my account in mysudoers  rna ALL = NOPASSWD: ~/scripts/getLog save &amp; exit chmod  sudo chmod 0440 /etc/sudoers.d/mysudoers   Error log when I type sudo :  rna@rna-P580:/etc/sudoers.d$ sudo                                                                sudo: &gt;&gt;&gt; /etc/sudoers.d/mysudoers: syntax error near line 0 &lt;&lt;&lt; sudo: parse error in /etc/sudoers.d/mysudoers near line 0 sudo: no valid sudoers sources found, quitting sudo: unable to initialize policy plugin
How to modify an invalid '/etc/sudoers' file? How do I edit an invalid sudoers file? It throws the below error and it's not allowing me to edit again to fix it.  Here is what happens:  $ sudo visudo &gt;&gt;&gt; /etc/sudoers: syntax error near line 28 &lt;&lt;&lt; sudo: parse error in /etc/sudoers near line 28 sudo: no valid sudoers sources found, quitting
Editing old questions Remove the extra headers from the e-mails sent Can Excel show a formula and its result simultaneously? Expanding GCDs Reasoning: $\gcd(a,c) \cdot \gcd(b,c) = \gcd(ab,bc,ac,cc)$ I'm just wondering the reason behind being able to do this:  $$\gcd(a,c) \cdot \gcd(b,c) = \gcd(ab,bc,ac,cc)$$  Is there an elementary proof to the reasoning behind this? $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution. What is the value of $p^2q + q^2r + r^2p$ for the given cubic equation? If p,q,r are the real roots of the equation $x^3-6x^2+3x+1=0$ then find the value of $p^2q + q^2r + r^2p$?  My Attempt: I tried $(p+q+r)(pq+qr+rp)$ but couldn't really figure out what to do with the extra terms. The roots are also not trivial to find. Any help would be appreciated. Properly discontinuous action: equivalent definitions Let us define a properly discontinuous action of a group $G$ on a topological space $X$ as an action such that every $x \in X$ has a neighborhood $U$ such that $gU \cap U \neq \emptyset$ implies $g = e$. I would like to prove that this property is equivalent to, having given $G$ the discrete topology and in the $X$ locally compact Hausdorff case, the map $G \times X \rightarrow X \times X$ given by $(g, x) \mapsto (x, gx)$ being proper (i.e. closed and preimage of compact sets is compact) plus the action being free.  I have managed to prove one direction, that is, if the action is proper and free with $G$ having the discrete topology then it is properly discontinuous. I'm having trouble though with the other direction. Here is an attempt: let's denote by $\rho : G \times X \rightarrow X \times X$ the map $\rho(g, x) = (x, gx)$. Suppose $K \subset X \times X$ is compact. We wish to show $\rho^{-1}(K)$ is compact. Let $(g_i, x_i)$ be a net in $\rho^{-1}(K)$. Then $\rho(g_i, x_i) = (x_i, g_i x_i)$ admits a convergent subnet, so passing to it we may assume $x_i \rightarrow x$ and $g_i x_i \rightarrow y$. Essentially we must now find a way to prove $g_i$ converges, but I can't seem to do this. Any hints? Prove that if $A$ is transitive, then $\mathcal{P}(A)$ is transitive too. My book on set theory says     Prove that is a set $A$ is transitive, then $\mathcal{P}(A)$ is also transitive, where $\mathcal{P}(A)$ is the power set of $A$.   I don't know understand how. Let $A=\{a,b\}$. Clearly it is transitive. Now $\mathcal{P}(A)=\{\{a\},\{b\},\{a,b\},\emptyset\}$. This is clearly not transitive. Proof that a sum of Bernoulli rvs has Binomial distribution Let $\{X_n, n \geq 1\}$ be iid Bernoulli random variables with $\Pr(X_1 = 1) = p = 1 - \Pr(X_1 = 0)$ and let $S_n = \sum\limits_{i=1}^n X_i$ be the number of successes in $n$ trials. Show $S_n$ has a binomial distribution by the following methods:  1.) Prove for $n\geq0,$ $1\leq{k}\leq{n+1}$  $$\Pr(S_{n+1} = k) = p\Pr(S_n = k - 1) + q\Pr(S_n = k)\text{ where }q = 1-p.$$  2.) Solve the recursion using generating functions.  For the first method, I think I need to use convolution of $n + 1$ Bernoulli rvs, but I cannot immediately see the progression. For the second method, I don't exactly understand how to proceed either. This is problem 1.2 from Resnick's Adventures in Stochastic Processes. How can I reorder my divs using only CSS? Given a template where the HTML cannot be modified because of other requirements, how is it possible to display (rearrange) a div above another div when they are not in that order in the HTML? Both divs contain data that varies in height and width.  &lt;div id="wrapper"&gt;     &lt;div id="firstDiv"&gt;         Content to be below in this situation     &lt;/div&gt;     &lt;div id="secondDiv"&gt;         Content to be above in this situation     &lt;/div&gt; &lt;/div&gt; Other elements   Hopefully it is obvious that the desired result is:  Content to be above in this situation Content to be below in this situation Other elements   When the dimensions are fixed it easy to position them where needed, but I need some ideas for when the content is variable. For the sake of this scenario, please just consider the width to be 100% on both.  I am specifically looking for a CSS-only solution (and it will probably have to be met with other solutions if that doesn't pan out).   There are other elements following this. A good suggestion was mentioned given the limited scenario I demonstrated&mdash;given that it might be the best answer, but I am looking to also make sure elements following this aren't impacted. I have a MacBook Pro 2009 running MacOS El Capitan. Recently the Wi-Fi has stopped working.   When I click on the Wi-Fi icon it displays the message "No hardware installed", but Ethernet still works.   Is there any way of fixing this by using a Wi-Fi dongle or something of the sort, if so do I need a specific dongle and how do I install it? Subgradients of non-convex functions In  (section 2.3), it is stated that:     A point $x^*$ is a minimizer of a function $f$ (not necessarily convex) if and only if $f$ is subdifferentiable at $x^*$ and $0 \in\partial f(x^*).$   Could anybody provide me with references for a proof of the above statement?   Is there a reference where we can learn more about subgradients of non-convex functions?  In Section 3. Calculus of subgradients of the above notes,many properties of subgradients are presented for convex functions. I would like to know which ones among these properties still hold for non-convex functions.  Thank you in advance! $\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of the the real numbers.    My question is, is it possible for all $\aleph_1$ subsets of $\Bbb{R}$ to have Lebesgue measure $0$?  This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd.  But is $\mathsf{ZFC}$ + $\lnot\mathsf{CH}$ + "all subset of $\Bbb{R}$ of cardinality $\aleph_1$ have Lebesgue measure $0$" consistent?  If not, what if we replaced Lebesgue measure with some other measure?  It may be worth noting that, although there may be subsets of $\Bbb{R}$ with cardinality $\aleph_1$, there are no subsets of $\Bbb{R}$ which have the order-type $\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\Bbb{R}$.  Any help would be greatly appreciated.  Thank you in Advance. word choice - realize notice find out figure out I am interested in finding out how confident my model (say Logistic Regression) is in predicting the label of a new data point. For example, if it is not confident, I better abstain from making a prediction.   Logistic Regression outputs probabilities, which gives you a notion that there is confidence in prediction. But in fact, it is not. $P(y|x)$ being 0.3, doesn't tell me that the model is confident in its prediction or not. All, we can say is that it believes 100% that class A is 0.3 and other class is 0.7.  Can we use confidence intervals of LR as some sort of confidence in prediction? E.g., larger the difference between upper and lower bound for a particular data point, less confident it is, and vice-versa?  If not, how can we build confidence in our prediction? Can anybody guide me to some paper or field of study? Disposing WPF User Controls Origins of the nose-bleed in anime/manga In a similar spirit to my , I wonder who was the first artist to feature a nose-bleed when thinking, shall we say, "provocative thoughts"?  Did it become popularized in anime/manga because of the popularity of the artist's work or because it is a common belief in japanese folklore? Why does $\lim_{n\to\infty} z_n=A$ imply $\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A$? I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you. Reset post score to 0 on migration C99 stdint.h header and MS Visual Studio