Recently, I have been reading a book about TOEFL grammar. I came across this multiple-choice problem: I have _______ money left. (A) a great number (B) so small (C) only a little (D) only a few The answer given is (C). I don't understand why (B) is not the correct answer. Does anyone know the reason? Thank you very much.
Is there a difference between them? If so, how and when are they used?  For example:     I fixed a little/small typo.
I want to change my field from molecular biophysics to theoretical ecology. While they are basically required some same basic background (physics, math, programming), the rest is quite irrelevant. In my current draft of this part, about a half is about this the problem and the result of the work, another half is for my contribution and what I've learnt. Having said that, should I focus less on describing the work and leave space for the thing that will be useful for the next research?  Below is copy from .     What have you already done? What problems have you solved, or at least worked on? What independent projects have you been part of? What were your key contributions? What did you learn? What did you teach the world? How do your results compare to what was already known? What original ideas are you most proud of? Be specific, technical, credible, and confident (but not arrogant). Refer the reader to your web page for more details. Have a web page with more details: preprints, project reports, source code, videos, etc. Divergence of $\vec{F} = \frac{\hat{\mathrm{r}}}{r^{2}}$ Consider the vector field $$\vec{\mathrm{F}} = \frac{\hat{\mathrm{r}}}{r^{2}},$$ then the divergence of this field is: $$\vec{\nabla}\cdot\left(\frac{\hat{\mathrm{r}}}{r^{2}}\right) = 4\pi\delta^{3}(\mathrm{\vec{r}})$$  What is the proof of this relation? Comparing nullable column values using "IS NULL" versus "= NULL" I'm dealing with nullable values in a column of a table, i'm wondering why this:  with cte as (     select 'c1'::text as c1,null::text as c2     union     select 'c2'::text as c1,null::text as c2 ) select * from cte where c2 is null   returns a result set with the two "rows" having a null value in c2; comparting to this query:  with cte as (     select 'c1'::text as c1,null::text as c2     union     select 'c2'::text as c1,null::text as c2 ) select * from cte where c2 = null   that returns an empty result set!!  so it looks like that field is null is actually different from field = null? the question is... is there "null-compatible operator" to use? so i can write the comparision like: column_name = $1 and argument $1 can be a not null or null value. IF $1 is a null value then it will semantically equals to write as IS NULL, ELSE (if $1 is not a null value) then it would be semantically equals to = $1.  P.D: This question also applies to the case is NOT NULL versus &lt;&gt; NULL I'm trying to find a copy of 6.20 to run a diff on a client's Drupal install. I believe there's been changes to core.   How can I get version 6.20? Additionally, how can I download any older Drupal versions? Are DSLRs a dying breed, making now the time to switch to a mirrorless camera system? Can I use javascript to dynamically change a video's source? Proving orthogonality of vectors in a complex vector space I'm taking a course in Real Analysis and have come across the following question which I initially thought looked quite simple:  Let $X$ be an inner product space over $\mathbb{K}$ ($\mathbb{K} = \mathbb{C}$ or $\mathbb{R}$) Show that x and y are orthogonal if and only if $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$ for all $\mu, \lambda \in \mathbb{K}$.  I have proven the statement in one direction (starting with orthogonality and ending up with $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$). However, I seem to be getting stuck trying to go in the other direction. The following is what I have so far:  $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + 2\bar{\lambda}\mu Re&lt;x,y&gt; + ||\mu y||^2$  Then we must have 2$\bar{\lambda}\mu Re&lt;x,y&gt; = 0$ for the desired equality to hold, and so $Re&lt;x,y&gt; = 0$.   However, for orthogonality I want $&lt;x,y&gt; = 0$ and can't see how I can make this next step.  Any help would be greatly appreciated. How to put a brace on the right, not left, to group cases? Proving that any permutation in $S_n$ can be written as a product of disjoint cycles I have attempted a proof of this, but upon looking at my notes, I feel it might be incorrect: it is noticeably simpler than the one in my notes.  Proposition: any permutation in $S_n$ can be written as a product of disjoint cycles.  Proof (attempt): $s\in S_n$. Let $\langle s\rangle$ act on $X=\{1,\cdots,n\}$. Define $x\sim x'\iff x'\in\text{Orb}(x)$ for $x,x'\in X:$ this is clearly an equivalent relation, hence the orbits partition $X$. Each orbit corresponds a cycle in $s$, so we are done.  Does this hold, or have I missed something vital? Fitting unbreakable large equation on one line in a twocolumn layout After booting I ran systemd-analyze blame and here are the results:       21.596s systemd-journal-flush.service      18.658s dev-sda8.device      15.099s dev-loop33.device      15.034s dev-loop19.device      15.012s dev-loop34.device      14.989s dev-loop21.device      14.877s dev-loop15.device      14.866s dev-loop26.device      14.773s dev-loop27.device      14.684s dev-loop30.device      14.677s dev-loop32.device      14.649s dev-loop35.device      14.590s dev-loop25.device      14.267s dev-loop23.device      14.192s dev-loop24.device      14.156s dev-loop29.device      14.133s dev-loop16.device      14.065s dev-loop31.device      14.059s dev-loop28.device      13.821s dev-loop20.device      13.531s dev-loop22.device      13.495s dev-loop14.device      13.364s dev-loop18.device   What are these dev-loopxx.device (xx denotes numbers) services and why are they taking so much time? Are they related to the mounting of snaps? Can I reduce the boot time by disabling them? I'm running Ubuntu 18.04 alongside Windows 10. Why are C character literals ints instead of chars? In C++, sizeof('a') == sizeof(char) == 1. This makes intuitive sense, since 'a' is a character literal, and sizeof(char) == 1 as defined by the standard.  In C however, sizeof('a') == sizeof(int). That is, it appears that C character literals are actually integers. Does anyone know why? I can find plenty of mentions of this C quirk but no explanation for why it exists. Talk to him is what I did Talk to him is what I did.  (this sentence is from )  "Talk" is a verb here that is not in the imperative mood. I suppose, it's in the infinitive form. But then how can it stand alone, without "to"?  What will be if we add "to" to "talk"?  To talk to him is what I did.  If we can do so, then what will difference be between the initial sentence and the new one with "to"?  Maybe, there are some omitted words, after inserting which the sentence will immediately become understandable? I mean, maybe, there is an ellipsis?  Thanks!  To make it clearer:  Upd.1: if we type "bare infinitive" in google, we can't find any sites that could explain the absence from "to" before "talk".  Upd.2: Why can we replace "to talk" with only "talk" in "To talk to him is what I did."  Upd.3: I have my own logic but I don't know whether it's right or not:  1) I think that without an ellipsis "Talk to him is what I did." is to look like "I did talk to him is what I did do."  2) Also I know that we can't use "to" before "read" in "What we must do is read the manual." because in the left part we don't have "to" before "do".  By this logic, we can't use "to" before "talk" in "Talk to him is what I did." because in "what I did do" we don't have "to" before "do".  Am I right or not? Getting the ID of the element that fired an event Is there any way to get the ID of the element that fires an event?  I'm thinking something like:    $(document).ready(function() {   $("a").click(function() {     var test = caller.id;     alert(test.val());   }); }); &lt;script type="text/javascript" src="starterkit/jquery.js"&gt;&lt;/script&gt;  &lt;form class="item" id="aaa"&gt;   &lt;input class="title"&gt;&lt;/input&gt; &lt;/form&gt; &lt;form class="item" id="bbb"&gt;   &lt;input class="title"&gt;&lt;/input&gt; &lt;/form&gt;     Except of course that the var test should contain the id "aaa", if the event is fired from the first form, and "bbb", if the event is fired from the second form. Let $X_1, X_2, \ldots X_n$ be random variables with mean $\mu$ and variance $\sigma^2$. We define the sample mean $\bar{X} = \sum_{i = 1}^n X_i$.  I am having trouble understanding why the variance of $\bar{X}$ is $\sigma^2 / n$.  Here is what I worked out thus far:  $$\begin{gather} \sigma_\bar{X}^2 = E((\bar{X} - \mu)^2) = E(\bar{X}^2 - 2\bar{X}\mu + \mu^2) = E(\bar{X}^2) - 2\mu E(\bar{X}) + E(\mu^2) = E(\bar{X}^2) - \mu. \end{gather}$$  However, I'm not sure what to do with the $E(\bar{X}^2)$. What are all the font styles I can use in math mode? As well as many of the other features talked about during macOS Mojave's release like dark mode and the addition of new apps like Home, another key feature of macOS Mojave seems to be it's heightened focus on both privacy and security.   As quoted under the "Camera and Microphone Now Require Your Permission" section of  article:      Apple also announced that other categories of data, such as your Messages history and Mail database, will be protected in a similar manner to macOS Mojave's new camera and microphone permissions.   I understand that after the update, when an app requests to use your camera or microphone, you will receive an alert like the one below:    However, I am more intrigued by the following:     ...other categories of data, such as your Messages history and Mail database, will be protected in a similar manner...   I have been unable to find much information about this online and I do not currently have the pre-release version of macOS Mojave installed and am therefore unable to test this new feature but I am curious to know what exactly this feature will mean.   Has Apple publicly documented how this will work? Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.   Suppose also that balls are drawn from the box one at a time at random.  What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.  The answer I was given is $\dfrac{7}{12}$ and a general equation is: $$ \dfrac{b g}{1-b}+\dfrac{b g}{1-g} $$ where $$ g=\dfrac{20}{60},b=\dfrac{30}{60} $$ but why? How can I respawn the Ender Dragon? I have a server and the Ender dragon has been defeated.  However, returning to The End does not respawn the Ender Dragon.  How can I get the final boss back on a vanilla server without using mods? GCD of rationals Disclaimer: I'm an engineer, not a mathematician  Somebody claimed that $\gcd$ only is applicable for integers, but it seems I'm perfectly able to apply it to rationals also:    $$ \gcd\left(\frac{13}{6}, \frac{3}{4} \right) = \frac{1}{12} $$  I can do this for a number of cases on sight, but I need a method. I tried Euclid's algorithm, but I'm not sure about the end condition: a remainder of 0 doesn't seem to work here.    So I tried the following:    $$\gcd\left(\frac{a}{b}, \frac{c}{d} \right) = \frac{\gcd(a\cdot d, c \cdot b)}{b \cdot d}$$  This seems to work, but I'm just following my intuition, and I would like to know if this is a valid equation, maybe the correct method. (It is in any case consistent with $\gcd$ over the natural numbers.)  I'm not a mathematician, so please type slowly :-)