I want to download YouTube videos as mp3 files in its best quality. The below code helps to download youtube videos as m4a but not us mp3  youtube-dl -f bestaudio --audio-quality 0 --audio-format mp3 https://www.youtube.com/playlist?list=PLYRruMbyFRcBVdVN8v4FNkIKkXvL-bZn_   can you tell me how to use bestaudio parameter to download YouTube video as mp3
There are several Q&amp;A threads that explain how to download youtube videos using the terminal.  However, I would also like to learn how to extract the video's soundtracks as MP3 files--also using only the terminal.  Answers briefly explaining how to use youtube-dl or other similar utilities before explaining how to extract the MP3 would be ideal for the sake of having all the information in one place--even though this aspect has been covered in other posts.
Is there a difference between Bayesian and Classical sufficiency? The title pretty much says it all. I wonder whether there is any difference in the way Bayesians understand sufficiency vs. the way orthodox statistics understands sufficiency, or are they equivalent?  If there is a difference, what is it? I have a Storex eZee Tab 702 tablet. My Google Play Store was working very well until a few weeks ago, when it started saying,     Error while retrieving information from server [DF-DFERH-01]   I tried to clear the cache, clear the data, force stop, and downloading another Google Play Store, but I'm still having the same problem.  How can I fix this issue? Why is the derivative of a circle's area its perimeter (and similarly for spheres)? Does the review queue page show only reviews requiring your action? Does it differ, and when does it? This confusion came up in comments .  On the site I use most actively (RPG.SE), I've noticed that the counts for each review queue appear to only indicate the reviews available to me requiring my action. Just to be clear, I'm talking about this number here:        Since I'm a 10k user, that's pretty easy to tell: even after doing all my reviews, and the review page saying there's 0 items in each queue, I might still have that 10k notice in my top bar saying there's [5] reviews pending, or some other number. So they must still be there, just waiting on other users to finish them, and it just doesn't show them to me. This is fine, I like this behaviour.  However, on Meta SE about an hour ago, there was 1 item in the close vote review queue. I reviewed that item. The count stayed at 1 on the review page. When I clicked to re-enter the queue in case there were new items, I was presented with this:        Another user in that meta question I linked claims this is normal behaviour for them on their own site (Christianity.SE I think?).  What's going on here? Are the review queues supposed to be customised just to what I personally can actually review at the time? Is there a bug, are sites customised differently, or is there just some kind of caching thing leading to a time delay on the review count getting updated? (As they say, it is always caching, but is caching really the culprit here too?) Is a DNA molecule a single strand of polynucleotide or two of them linked together? Our molecular biology teacher told us that a double helix of DNA was composed of two DNA molecules linked together by hydrogen bonds. The thing is, until now, I always thought a DNA molecule was composed of two strands, those being polynucleotides, both of them being linked together. I can't find a link which is saying the same as my teacher, even if it seems technically correct to call a double helix a dimer of two DNA molecules.  I was curious to know what was the exact terminology. Proof that $\mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $ Where $ f: V \rightarrow V  $ is a linear operator on an inner product space $V$, I'm trying to show that $$ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $$  i.e. the Kernel of the adjoint is equal to the orthogonal complement of the image of f.  I taking the inner product of $f(v)$, where $v \in V$  and an arbitary element of the kernel, $k \in \mathrm{ker}(f^*) $  $$\langle f(v), k\rangle = \langle v, f^*(k)\rangle = \langle v, 0\rangle = 0$$  And then taking the inner product of $f(v)$ with $w$, where $w \in  \mathrm{im}(f)^{\perp}$  By definition  of the orthogonal complement to the image of f,   $$ \langle f(v), w\rangle = 0 $$  My question is, now that we have   $$ \langle f(v), k\rangle = \langle f(v), w\rangle = 0 $$  What can we conclude? I would like to conclude that because $w$ and $k$ are arbitary elements of their respective subspaces, that these must be equal, i.e. $ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $  However, I am not sure this is mathematically sound, is that conclusion valid or are there any technical barriers or additional statements that would have to be made to make that watertight? Prove that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw) \subset \mathbb{A}^4$, is not a unique factorization domain I want to show that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw)\subset\mathbb{A}^4$, is not a unique factorization domain.    Morally, all we need to do is find some nonzero element that can be written in two distinct ways as a product of irreducible elements, but nothing is coming to mind immediately. Thoughts? Even my main monitor is shown as unknown. I have tried setting up dual monitors using Nvidia settings with no joy however it does recognize that I have 2 monitors.  This was not an issue in 11.04. My problem with the Nvidia settings appears to be that my 2 monitors require different resolutions but even when I set them to a resolution that they both can handle, it is then only appearing in cloned mode. I really hope someone can help or I will have to roll back to 11.04.  Why won't Ubuntu 11.10 recognize that I have 2 Monitors?   Here is the contents of my xorg.conf:  # nvidia-settings: X configuration file generated by nvidia-settings # nvidia-settings:  version 280.13  (buildd@rothera)  Thu Aug 11 17:28:49 UTC 2011  Section "ServerLayout"     Identifier     "Layout0"     Screen      0  "Screen0" 0 0     InputDevice    "Keyboard0" "CoreKeyboard"     InputDevice    "Mouse0" "CorePointer"     Option         "Xinerama" "0" EndSection  Section "Files" EndSection  Section "InputDevice"     # generated from default     Identifier     "Mouse0"     Driver         "mouse"     Option         "Protocol" "auto"     Option         "Device" "/dev/psaux"     Option         "Emulate3Buttons" "no"     Option         "ZAxisMapping" "4 5" EndSection  Section "InputDevice"     # generated from default     Identifier     "Keyboard0"     Driver         "kbd" EndSection  Section "Monitor"     # HorizSync source: xconfig, VertRefresh source: xconfig     Identifier     "Monitor0"     VendorName     "Unknown"     ModelName      "CRT-0"     HorizSync       30.0 - 83.0     VertRefresh     50.0 - 76.0     Option         "DPMS" EndSection  Section "Device"     Identifier     "Device0"     Driver         "nvidia"     VendorName     "NVIDIA Corporation"     BoardName      "GeForce 7025 / nForce 630a" EndSection  Section "Screen"     Identifier     "Screen0"     Device         "Device0"     Monitor        "Monitor0"     DefaultDepth    24     Option         "TwinView" "1"     Option         "TwinViewXineramaInfoOrder" "CRT-0"     Option         "metamodes" "CRT: 1280x1024 +1024+0, DFP: 1024x768 +0+0"     SubSection     "Display"         Depth       24     EndSubSection EndSection"   ok I have tried disper -e. Problem is that Nvidia settings appear to be treating both monitors as the same one. so changing the settings to one has no effect, but change the other and it affects both screens identically.  Results of disper -l as follows:      carter@carter-desktop:~$ disper -l display DFP-0: HP vs17  resolutions: 320x240, 400x300, 416x312, 512x384, 576x432, 680x384, 640x480, 720x450, 640x512, 800x512, 840x525, 800x600, 960x540, 832x624, 960x600, 896x672, 1024x768, 1152x864, 1280x960, 1280x1024 display CRT-0: CRT-0  resolutions: 320x240, 400x300, 416x312, 512x384, 576x432, 680x384, 640x480, 720x450, 640x512, 800x512, 840x525, 800x600, 960x540, 832x624, 960x600, 896x672, 1024x768, 1152x864, 1360x768, 1280x960, 1440x900, 1280x1024, 1400x1050, 1600x1024, 1680x1050, 1600x1200, 1920x1080, 1920x1200, 1792x1344 carter@carter-desktop:~$ What is a Seaman's book, and can it really replace a passport? What is a "Seaman's book", and can it really replace a passport for some or all immigration purposes?  Even Wikipedia the all-knowing does not seem to know, although it contains many tantalizing references.  :     Seaman deserters may not have a passport but rather hold a Seaman's Book, (a national identity document issued to professional seamen that contains a record of their rank and service career).   :     Possession of one of the following documents is sufficient to demonstrate [right to abode] ...         Hong Kong Seaman's Identity Book Term to describe users' initial dislike of user interface change What is the best term to describe the fact that most users have an initial dislike of any change to a UI even though it may really be net better over time?  Note: Question originally appeared How is Cooper's age justified? I am trying to find the equation of characteristic curves and solution of the partial differential equation $$x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=u$$ $u(x,0)=\sin(\frac{\pi}{4}x)$ .   What I did: $$\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{u}$$ with  From first two equalities $$\frac{dx}{-y}=\frac{dy}{x}$$ by solving we get $x^2+y^2=c_1$ for some constant $c_1$. Now I am stuck with the third and rest of the equalities to come up with a suitable manipulation to solve the problem. How can I do this? Any help would be great. Thanks. R: Trouble using SMOTE package "invalid 'labels'" Is there really a way to make a true toon shader material in Cycles? Is there a way to make a true toon shader material in Cycles?  Here's, in order of importance, what would be ideal to find in such a shader:   Must use Cycles Must be an actual shader Must not rely on lighting Must be independent of camera position Must react directly to lighting rather than manually having to change it to emulate it Should be easily made into a node-group Should have the following configurability:   Main color Darkness color Shadow depth (how far the shadow goes from the unlit side)    Below, I'll be listing some already-found methods and their flaws:    Cycles's Toon BDSF:  Source: Blender  Result:    Node setup:    Cons:   Not a clear-cut two-color shading Reacts to diffusion of surrounding objects      Mask &amp; Ramp:  Source:   Result:    Node setup (Compositor):    Cons:   Very grainy Does not support colour textures Not easy to change shadow depth in a nodegroup Not a material, requires compositor Requires use of object indexes     Shadeless Fresnel:  Source:   Result:    Node setup:    Cons:   Uses camera's direction Ignores lighting Not easy to change shadow depth in a nodegroup     Shadeless Normal:  Source:   Result:    Node setup:    Cons:   Ignores lighting Direction of shadow is manually changed Not easy to change shadow depth in a nodegroup   However, this one is easily the best one. It's pretty great, and would work still images very well.    That's all I've got. Does anyone else have any methods that fall under most of the criteria I listed above? Also note that OSL is totally fine, if it works. Probability of complete coupon set in $K$ boxes where a box has $N$ distinct coupons? Find the area ratio between ABC and the hatched triangle. Could you help me?  Find the area ratio between ABC and the hatched triangle.    Thanks in advance. Intersection of all $p$-Sylow subgroups is normal Fastest way to determine if an integer's square root is an integer I'm looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer):    I've done it the easy way, by using the built-in  function, but I'm wondering if there is a way to do it faster by restricting yourself to integer-only domain. Maintaining a lookup table is impractical (since there are about 231.5 integers whose square is less than 263).   Here is the very simple and straightforward way I'm doing it now:  public final static boolean isPerfectSquare(long n) {   if (n &lt; 0)     return false;    long tst = (long)(Math.sqrt(n) + 0.5);   return tst*tst == n; }   Note: I'm using this function in many  problems.  So no one else will ever have to maintain this code.  And this kind of micro-optimization could actually make a difference, since part of the challenge is to do every algorithm in less than a minute, and this function will need to be called millions of times in some problems.    I've tried the different solutions to the problem:   After exhaustive testing, I found that adding 0.5 to the result of Math.sqrt() is not necessary, at least not on my machine. The  was faster, but it gave incorrect results for n >= 410881.  However, as suggested by , we can use the FISR hack for n &lt; 410881. Newton's method was a good bit slower than Math.sqrt().  This is probably because Math.sqrt() uses something similar to Newton's Method, but implemented in the hardware so it's much faster than in Java.  Also, Newton's Method still required use of doubles. A modified Newton's method, which used a few tricks so that only integer math was involved, required some hacks to avoid overflow (I want this function to work with all positive 64-bit signed integers), and it was still slower than Math.sqrt(). Binary chop was even slower.  This makes sense because the binary chop will on average require 16 passes to find the square root of a 64-bit number. According to John's tests, using or statements is faster in C++ than using a switch, but in Java and C# there appears to be no difference between or and switch. I also tried making a lookup table (as a private static array of 64 boolean values).  Then instead of either switch or or statement, I would just say if(lookup[(int)(n&amp;0x3F)]) { test } else return false;.  To my surprise, this was (just slightly) slower. This is because . How can I tell if a CVE has been fixed in Ubuntu's repositories? How to move Ubuntu to an SSD