I made a rough calculation (probably having made a miscalculation though) following  and  to find out the number of air molecules in 1 cubic centimetre of air.   The first website says there are about $0.02504\times10^{27}$ air molecules in 1 cubic metre. The the second one says the diameter of an air molecule is about $4\times 10^{-10}$m. Then the area of the molecule is about $4\pi\times 10^{-20} \approx 12\times 10^{-20}$.  Then we get the number of air molecules in one cubic metre to be about $12\times0.025\times10^7$, and in one cubic centimetre to be about $12\times0.025\times10^5 = 30000$.  So if there are 30000 air molecules in 1 cubic centimetre, it's almost impossible for light to miss hitting all of them. Following this logic, I can't comprehend how light propagates in air (I would expect it to be scattered, absorbed etc). How does physics explain this?   Furthermore, since the speed of light in a vacuum is constant, but light is "slower" in higher refractive index materials, can I assume light travels a longer path when it travels through a medium with a refractive index higher than empty space? If not, what decreases its speed?   (I thought two questions are related as they both question how light travel in air.)
One of the great 'coincidences' in physics is that the Sun happens to shine most brightly at exactly the wavelengths our eyes can see; it's an easy explanation that our eyes evolved to make the most out of what was available.  However, this spectral window is also , with Rayleigh scattering and the ozone layer protecting us from the UV and strong water vapour absorption in the IR until way into the radio range.  My question is: is this a coincidence? Were there geological factors that made it so? Or is it just lucky chance? (I also notice other planets in our neighbourhood weren't quite so lucky.) Or had atmospheric absorption been different our eyes would have tuned to whatever spectral window were brighter as a whole?
How does the subject-verb agreement work when a singular noun describes a plural one? Of the following, which sentence is righter?  The computers is a good choice. The computers are a good choice.  Is there a rule about these types of formations? Please explain. We currently have some indexes where ALLOW_PAGELOCKS is set to off. This presumably was done in order to reduce deadlocks. However I doubt that it would really had an effect back then.   Now I am trying to understand WHEN SQL Server actually chooses to start locking pages rather than keys in a clustered index. I asked Jonathan Keyhaisas recently and he told me that this could happen if I am touching rows on several subsequent pages. However I didn't manage to get any exclusive page locks by updating rows in a clustered index with a sample query.   Could you help me understand page locks better with a sample query and table? I am running SQL Server 2008 SP4.  Thanks in advance  Martin Where can I get Blender test builds? I know that  provides downloads for official releases. I see on forums that people use newer versions of Blender. Where can I get newer version of Blender (nightly builds, alpha versions... etc). Derive Variance of regression coefficient in simple linear regression A probability theory question about independent coin tosses by two players Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.    The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]&amp; = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}   How does one simplify this?      Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2. Determine if $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ converges Mis-aligned shapefiles in QGIS? What are point objects? Travel without an Italian residence permit in the Schengen area My ceiling fixture has all of a sudden stopped working properly. Half the diodes are dark and the others are blinking. Oddly, the same exact ones on each strip work and don't work. I have included a picture to show which ones are blinking and which ones don't work. I am wondering if it is something that I can fix? Maybe replace that driver? And if so, is there a good guide online to walk me through that process? Thanks in advance. I have a folder on my desktop which is named ".." (which itself contains subfolders with an empty name). The folder can't be deleted the usual ways, even not when using the CLI. Does anyone know how I can remove this folder on Windows Vista 64-bit?  Renaming does not work, either (obviously).  This is what "dir /x" echoes:  09.10.2009  15:04    &lt;DIR&gt;                       . 09.10.2009  15:04    &lt;DIR&gt;                       .. 08.03.2007  11:18    &lt;DIR&gt;                       cgi-bin                0 Datei(en),              0 Bytes                3 Verzeichnis(se), 45.866.037.248 Bytes frei Why don't brakes come with 'Safety Levers' any more? It used to be very common to see bikes with dual pull brake levers (if that's even the right name) as shown below. However now it seems almost impossible to find brakes like this.  Even finding a picture to post was hard.  It seems like it would be quite useful to be able to brake from various hand positions.  There was an answer I read a while back (can't find it now) that stated that there was a reason they got rid of those, but no explanation as to why. So any technical reason, or is it just for looks? Honey changing into unusable, hardened clump I buy raw, unfiltered honey from the grocery store in jars sold/packed directly by the farmer. When I had used the last jar to about half, I noticed the honey beginning to dry(crystalize?) to the point where I couldn't use the honey around these spots.  Not too long after this, the entire jar had crystallized into a block of unusable hard stuff.   I didn't do anything differently with this jar than the last jar, but only now am I having the issue. My question(s):   Why is the honey crystallizing, and why so rapidly?   1) Is the honey unusable at this point?  2) Should I store the honey differently after purchasing it(move from jar to other container)?  3) I've never seen this happen to the filtered, strained(fake) honey from the store, so what causes this to happen in the raw and unfiltered form? Permalink rewrite with custom post type and custom taxonomy I think thats a quick one:  I've got:   Custom post type: Food  Custom taxonomy (Registered to Food): Fruits Terms in Fruits: Apple, Orange, Cherry   If I type in example.com/food?fruits=Apple,Cherry, I get all posts in Fruits with the terms Apple and Cherry, thats great!  But I would like to type in example.com/food/fruits/Apple,Cherry or example.com/food/Apple,Cherry to get the same result.  I´ve tried different permalink and rewrite plugins related to custom posts but nothing helped.  Thank you,  David Suppose we have $x^2+4x+3=0$  We differentiate both sides to get $2x+4=0$.  So $x=-2$  But $(-2)^2+4(-2)+3=0$   $4-8+3=-1$ is not equal to $0$.  Which tells that we cannot differentiate equalities. But in many proofs, like in integral calculus, differentiating both sides occurs. So, where is the fallacy? For all $1 \leq i < j \leq k$, the subtrees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$. Can someone please verify my proof or offer suggestions for improvement? I am aware that there is a similar question elsewhere, but I want help with my proof in particular.     Let $T_1, \ldots, T_k$ be subtrees of a tree $T$ such that for all $1 \leq i &lt; j \leq k$, the trees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$.   Suppose, for the sake of contradiction, that for every vertex $v \in V(T)$, there is a subtree $T'$ in the list $T_1, T_2, \ldots, T_k$ such that $v \notin V(T')$.  Pick $v_1 \in T_1$. There is a subtree $T^{(1)}$ such that $v_1 \notin T^{(1)}$. $T_1$ and $T^{(1)}$ have a vertex $y_1$ in common. Now, there exists a tree $T^{(2)}$ such that $y_1 \notin T^{(2)}$. But now, $T^{(2)}$ and $T^{(1)}$ have a vertex $y_2$ in common. Also, $T_1$ and $T^{(2)}$ have a vertex $y_3$ in common.  There exists a path $p_1$ in $T_1$ from $y_3$ to $y_1$. Also, there is a path $p_2$ in $T^{(1)}$ from $y_1$ to $y_2$ and a path $p_3$ in $T^{(2)}$ from $y_2$ to $y_3$. Note that the path formed by the union of $p_1$ and $p_2$ passes through $y_1$, while $p_3$ does not, since $y_1 \notin T^{(2)}$. Therefore, the union of $p_1, p_2$, and $p_3$ forms a cycle, contradicting the fact that $T$ is a tree. Therefore, it must be the case that there is a vertex $x$ in all of the $T_i$. How do you check if a selector matches something in jQuery? In Mootools, I'd just run if ($('target')) { ... }.  Does if ($('#target')) { ... } in jQuery work the same way? When I read about , I see that they should appear as solution to prevent  and  but when I actually try to write a code which would use Lightning promises it still doesn't look like synchronous code, it still resembles both pyramid of doom and callback hell.  main: function(component, helper) {     helper.promise1(component, helper).then(         $A.getCallback(function(record) {             return helper.promise2(component, helper, record).then(                 $A.getCallback(function(){                     console.log("Success");                     helper.showToastMessage(component, "Great Success!!!!")                 }),                  $A.getCallback(function(errorMessage){                     console.log("Rejected by second promise2");                     helper.showToastMessage(component, "Failure " + errorMessage);                 })              )         }),          $A.getCallback(function(errors){             console.log("Rejected by first promise1");             let message = errors &amp;&amp; Array.isArray(errors) &amp;&amp; errors.length &gt; 0 ? errors[0].message :'Unknown error';             helper.showToastMessage(component, "Failure " + message);         })      ) }   Is that possible to refactor this code to look more like synchronous code and without callback hell or pyramid of doom? Git ignore file for Xcode projects Which files should I include in .gitignore when using Git in conjunction with Xcode? Is it possible for test error to be lower than training error Is it possible to have test error lower than training error?  I have a classification problem with 2000 samples, 500 of which are positives, 1500 are negatives. I split my data into 70% training data, 30% test data.  Run random forest with 200 estimators and cv=10. I did this several times and compared the recall and precision score and notice the scores for my test set is significantly better. Is this possible?