How to accurately, unambiguously and concisely say something in the following cases:  Case 1. The predictor is significant, with 1.5615 times as much likely to get higher scores when it is true. Case 2. Subjects with A are 1.5615 times as much likely to get higher scores as those without A.  Maybe can I use &quot;chance&quot; instead of &quot;likely&quot; to form the sentence better?
Suppose John has 5 sweets. Is there any difference between the following two sentences?     Jack has 3 times as many sweets as John.      Jack has 3 times more sweets than John.   I prefer the first construction and would know unambiguously that Jack has 15 sweets in this case. However in the second construction I would be inclined to think that Jack has 20 sweets, since it seems to suggest 15 sweets in addition to the original 5.
How to detect the Internet connection is offline in JavaScript? Already I have lost all hope of looking at late answers etc. It's just a parade of meh. I can't make any kind of intelligent decision because I can't see the other answers to the question, for instance. There's no Looks Good, only Skip, and those feel different to me. So I've been more interested in .  But people keep approving crap while I'm looking at it. Two or three "the post was already approved; please visit to edit" and I'm just "screw this whole thing".  Can we do something like one reject bounces all existing votes: you need three approves in a row? Because a lot of people are stupidly approving and actually driving me away from the queue.    Note this is less about letting bad edits through and more about keeping reviewers motivated. Using same modulus for RSA Review sections to be always in the same place I find it not comfortable to have the order of the review sections changing all the time.  Within few seconds it changed from        to        Wouldn't it be better to have all of them in the same position, no matter the number of posts in each queue?  It assumes that the most important part is to lower the queue, although it is said there is no problem in having Close Votes with thousands of items. IMOH, most of us prefer going to one or other section (for example I like "Suggested edits") rather than to the queue with the most items, so having them changing their position makes it not that intuitive.  What do you think? Neatest proof that set of finite subsets is countable? I am looking for a beautiful way of showing the following basic result in elementary set theory:     If $A$ is a countable set then the set of finite subsets of $A$ is   countable.   I proved it as follows but my proof is somewhat fugly so I was wondering if there is a neat way of showing it:  Let $|A| \le \aleph_0$. If $A$ is finite then $P(A)$ is finite and hence countable. If $|A| = \aleph_0$ then there is a bijection $A \to \omega$ so that we may assume that we are talking about finite subsets of $\omega$ from now on. Define a map $\varphi: [A]^{&lt;\aleph_0} \to (0,1) \cap \mathbb Q$ as $B \mapsto \sum_{n \in \omega} \frac{\chi_B (n)}{2^n}$. Then $\varphi$ is injective hence the claim follows. (The proof of which is what is the core-fugly part of the proof and I omit it.). bash if -a vs -e option Both about -a and -e options in  is said:  -a file     True if file exists. -e file     True if file exists.    Trying to get what the difference is I ran the following script:  resin_dir=/Test/Resin_wheleph/Results  if [ -e ${resin_dir} ] ; then     echo "-e "; fi  if [ ! -e ${resin_dir} ] ; then     echo "! -e"; fi  if [ -a ${resin_dir} ] ; then     echo "-a"; fi  if [ ! -a ${resin_dir} ] ; then     echo "! -a"; fi   /Test/Resin_wheleph/Results exists and is a directory. And this is what I get:  -e -a ! -a   which seems to be a little strange (notice -a and  ! -a). But when I use double brackets (e. g. if [[ -e ${resin_dir} ]]) in the similar script it gives reasonable output:  -e -a   So:    What is a difference between -a and -e options? Why -a produces a strange result when used inside single brackets? Comparing signatures of apks I have two .apk files that are supposed to be different versions of the same app. What is the best way to check if they where both signed with the same key?  I want to make sure no mistakes were done and the newer version would be treated as an update of the same app. What is the difference between Affection and Friendship, if any? I've risen Pokemon Amie Affection to 5 hearts on my Riolo in the hopes that it will evolve as soon as it levels up, but I'm not certain that this is the same thing as friendship, and I'd like to not waste a level.  My question is this: What is the difference between Pokemon Amie Affection and the Friendship stat? $\operatorname{Hom}(X \times Z, Y) \cong \operatorname{Hom}(X, \operatorname{Map}(Z, Y))$ is not true in $\textbf{Top}$ I read that in the category of topological space it is not true that $\operatorname{Hom}(X \times Z, Y) \cong \operatorname{Hom}(X, \operatorname{Map}(Z, Y))$ for every $X, Y, Z$ topological spaces. In particular, I read that this does not work when $Z = \mathbb{R} \setminus \{ 1/n \mid n \in \mathbb{N}\}$. This was not explained, i.e. there was no example.   I am looking for an example or – maybe better – just a hint.  EDIT   $\operatorname{Map}(Z, Y)$ is endowed with the compact-open topology $\cong$ means just bijection; i.e. there is already an obvious bijection as sets, but the claim is that (I think) sometimes the "output function" is not continuous How many 6-letter permutations can be formed using only the letters of the word, MISSISSIPPI? I understand the trivial case where there are no repeating letters in the word (for arranging smaller words) but I stumble when this isn't satisfied. I'm wondering if there's a simpler way to computing all the possible permutations, as an alternative to manually calculating all the possible 6-letter combinations and summing the corresponding permutations of each. In addition, is there a method to generalize the result based on any P number of letters in a set with repeating letters, to calculate the number of n-letter permutations?  Sorry if this question (or similar) has been asked before, but if it has, please link me to the relevant page. I also apologize if the explanation is unclear, or if I've incorrectly used any terminology (I'm only a high-school student.) If so, please comment. :-) Permutations on word $MISSISSIPPI$. In how many ways can the letters of the word $MISSISSIPPI$ be rearranged ?   I am confused on whether it is  $\dfrac{11!}{4!4!2!}$ or $\dfrac{11!}{4!4!2!}-1$  since it is given rearranged and not arranged. Difference between Hashing a Password and Encrypting it The current top-voted to  states:     Another one that's not so much a security issue, although it is security-related, is complete and abject failure to grok the difference between hashing a password and encrypting it. Most commonly found in code where the programmer is trying to provide unsafe "Remind me of my password" functionality.   What exactly is this difference? I was always under the impression that hashing was a form of encryption. What is the unsafe functionality the poster is referring to? I just installed Solaris 11.3 on the laptop. This laptop have Broadcom BCM4312 WiFi card.  It looks like the install does not have a driver for this card just like 11.1 version.  I have a Linux installed on this same machine (Gentoo) and this OS does have a driver.  Is it possible to get the driver for this card somewhere? Maybe broadcom.com? Failing that is it possible to use Linux driver inside Solaris?  On the kind of related side - I tried to connect a wire and run Driver Manager, but it does not want to finish. All controls are  still disabled. Any idea?  Thank you for any pointers you can provide. My textbook explains the proof, which I don't understand:  "Consider two bases $v_1...v_p$ and $w_1...w_q$ of V. Since the vectors $v$ are linearly independent and the vectors $w$ span V..."  How exactly does $w$ span V?   The book then says the same for vectors $v$, that $v$ spans V and hence $p=q$, but I don't really understand how you can assume that given two sets of vectors that are basis, one set must span V. How do I see the type of a variable whether it is unsigned 32 bit, signed 16 bit, etc.?  How do I view it? Markov chains, cellular automata, and raster algebra I need to build a probabilistic cellular automaton for simulating land-use changes and am wondering how to implement this in a GIS. I understand that Markov chains are a way of doing this, and IDRISI has some sort of module for Markov chains, but I have no access to IDRISI.  that, with raster data, this can, in theory, be done with any GIS that can do raster algebra. Presumably this means that it can even be done in ArcGIS, and I think I've figured out how to build cellular automata in ModelBuilder (once I know with certainty how to make iteration work there).  Still, I can find next to no practical advice on this. Once I have two datasets, one showing the land use at a certain point in time (time A) and the other at a later time (time B), I imagine doing the following (correct me if I'm wrong):   If there are (for example) 3 types of land use, then there are 9 possible things that can happen to a cell from time A to time B; thus, reclassify each dataset such that each land use is represented by a different number (for example, 9, 11, and 14 at time A and 3, 4, and 5 at time B); Divide one raster by the other, obtaining a raster with 9 possible cell values. Determine the percentage of cells in the new raster that has each of the 9 values (I forgot, how does one do this in ArcGIS?); those will be the probabilities of each type of change (or lack thereof) in land use.   What I cannot figure out, though, is how to apply these probabilities during the next iteration. Finally, I wonder if there is a way of making an external factor (say, proximity to roads) influence the spatial distribution (not the overall likelihood) of change in my model.  Can anyone recommend where to find information on how to do this in ArcGIS? How to remove Apple System Apps How can I delete Apple OS X application such as Photos, iPhoto, and others that I never use? Proving $6^n - 1$ is always divisible by $5$ by induction I'm trying to prove the following, but can't seem to understand it. Can somebody help?     Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.   What I've done:  Base Case: $n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.  Assume true for $n = k$, where $k \geq 1$: $6^k - 1 = 5P$.  Should be true for $n = k + 1$  $6^{k + 1} - 1 = 5Q$  $= 6 \cdot 6^k - 1$  However, I am unsure on where to go from here. After lethal first strike damage to an indestructible creature, will it still do damage? If I were to attack with a 4/4 indestructible creature and it's blocked by a 4/2 creature with first strike, would my indestructible creature still attack during the second combat damage step, or would it just tap, not assigning any damage? Rigify generated rig, hand IK controller has unexpected rotation