Google says one can use the following method to remove a page from the search results: by putting -http://www.example.com/outdated.html into the sitemap. But what if the site does not exist anymore.
Before asking my question, I'll use an example to clarify the idea.  Say that a company used to sell horses and then they turn into a sheeps business. While some of the old pages can be mapped to the new pages using a 301 redirect, because they're really related in the content and the company don't want to lose users and link juice, there are a lot of pages strictly related to horses that they want to shutdown.  What's the best practice? Is there a way to say to users and Search Engines the content you were looking for won't be available any more?  Basically a 301 redirect of the closed pages to new ones seems not so good, because you'll show a content totally unrelated to the searched one.  P.S: I'm aware that probably the best fallback is to simply map those pages to the homepage but I hope in a smarter solution.
Ubuntu 18.04 can't resume after hibernate Hibernate worked correctly when I had been using Ubuntu 17.04. After upgrade to 17.10 I couldn't resume because of black screen after resume ().   After upgrade to 18.04 if I try to start my computer after hibernation it boots as if there was no hibernation done. Is there i.i.d. assumption on logistic regression? Is there i.i.d. assumption on the response variable of logistic regression?  For example, suppose we have $1000$ data points. It seems the response $Y_i$ is coming from a Bernoulli distribution with $p_i=\text{logit}^{-1}(\beta_0+\beta_1 x_i)$. Therefore, we should have $1000$ Bernoulli distributions, with different parameter $p$.  So, they are "independent", but are not "identical".  Am I right?    PS. I learned logistic regression from "machine learning" literature, where we optimize the objective function and check if it is good in testing data, without talking too much about assumptions.   My question started with this post  Where I try to learn more on statistical assumptions. Closing a window How to remove chromium-browser settings from Ubuntu 14.04? I have already gone through   and I'm really not getting why the solutions given there are not working for me.  What I'm doing is:   sudo apt-get purge chromium-browser   rm -rf ~/.config/chromium     Then reinstalling by running the following command:   sudo apt-get install chromium-browser    This must remove the settings of chromium-browser and do a fresh installation. The browser is uninstalling, but after being installed it is containing the previous settings, moreover it's not signing me out of even Gmail! How do people at Hogwarts usually tell the time? Classes and lessons at Hogwarts are presumably scheduled at fixed times of day. The first class might start, for example at nine in the morning.  To be on time for one's lessons, it is therefore crucial to know the current time but we know that electric watches of course do not work at Hogwarts because of all the magical interference. Are there magic watches? Is there maybe a spell to tell the time?  How do people at Hogwarts usually tell the time? How can I use Bengali script in an English document? The Cartesian product of a finite number of countable sets is countable The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i\}$$ Let $I_n=\{i \in \mathbb N \mid i \le n\}$ for $n\in \Bbb N$ and $A_i$ be countable for all $i\in I_n$. Then $\prod\limits_{i\in I_n}A_i$ is countable.   My attempt: Lemma 1: $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ and $\prod\limits_{i\in I_{n+1}}A_i$ are equinumerous for all $n \in \mathbb N$. (I presented a proof ) Lemma 2: If $A$ and $B$ are countable, then $A \times B$ is countable. (I presented a proof ) I will prove the theorem by induction on $n$. The theorem is trivially true for $n=0$. Assume that the theorem is true for $n=k$, then $\prod\limits_{i\in I_k}A_i$ is countable. Since $\prod\limits_{i\in I_k}A_i$ is countable (by inductive hypothesis) and $A_{k+1}$ is countable, then by Lemma 2 $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ is countable. Furthermore, by Lemma 1 $(\prod\limits_{i\in I_k}A_i) \times A_{k+1}$ and $\prod\limits_{i\in I_{k+1}}A_i$ are equinumerous, then $\prod\limits_{i\in I_{k+1}}A_i$ is countable too. Thus the theorem is true for $n=k+1$. This completes the proof.   Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help! 'Sorry' as a replacement for 'Excuse me'? Is it ok to say 'sorry' where you otherwise would use 'excuse me'?  Example: "Sorry, but could you show me the way to the next train station?" Excel: identifying if all values in a row are not equal For example an electron radiates when accelerated. So does a positron. But is the radiation emitted by accelerated positronium the sum of the radiation emitted by each separately? If not, why not? If so, does this provide a way of testing whether a given neutral particle is composite? For example, does a neutron bremsstrahlung when decelerated? VirtualBox error kernel headers I'm with Ubuntu 15.04 and I'm trying to create an emulator (genymotion), but when run virtualbox, i got the error:     Stopping VirtualBox kernel modules ...done.   Uninstalling old VirtualBox DKMS kernel modules ...done.   Trying to register the VirtualBox kernel modules using DKMSError! Your kernel headers for kernel 3.16.0-30-generic cannot be found.   Please install the linux-headers-3.16.0-30-generic package,   or use the --kernelsourcedir option to tell DKMS where it's located    ...failed!     (Failed, trying without DKMS)   Recompiling VirtualBox kernel modules ...failed!     (Look at /var/log/vbox-install.log to find out what went wrong)   I performed   uname -a   And the result is:     Linux douglas 3.16.0-30-generic #40-Ubuntu SMP Mon Jan 12 22:06:37 UTC 2015 x86_64 x86_64 x86_64 GNU/Linux   Someone help me?Already I uninstalled and installed again. Is the Stack Overflow source code available? ext4 partition size / free space discrepancies While creating a 250GiB backup partition for my data, I have noticed lots of discrepancies between reported partition size and free space in Nautilus, gParted, df, tune2fs, etc.  At first I thought it was a GiB / GB confusion. It was not.  Then I thought it could be ext4's reserved blocks. It was not.  Im completely puzzled. Here are some images. Here are the steps:   First, NTFS. 524288000 sectors x 512 bytes/sector = 268435456000 bytes = 268.4 GB = 250 GiB.      Nautilus say "Total Capacity: 250.0 GB" (even though its actually GiB, not GB). Apart from that minor mislabeling, so far, so good   Now, same partition, formated as ext4 with gparted:     First, Last and Total sectors are the same. It IS the same 250GiB partition. Used size is 4.11GiB (reserved blocks maybe?)    Nope. Looks like reserved blocks are 12.7 GiB (~5%. ouch!). But... why Total Capacity is now only 246.1 GiB ???. That difference (sort of) matches the 4.11 GiB reported by gparted. But... if its not from reserved blocks, what is it? And why gparted didnt report that 12.7GiB of used space?  $ df -h /dev/sda5 Filesystem            Size  Used Avail Use% Mounted on /dev/sda5             247G  188M  234G   1% /media/BACKUP   df matches Nautilus in reported free space. But.. only 188M used? Shouldnt it be ~12GB? And total capacity is still wrong. So i ran tune2fs to find some clues. (irrelevant output is ommited)  $ sudo tune2fs -l /dev/sda5 tune2fs 1.41.12 (17-May-2010) Filesystem volume name:   BACKUP Filesystem UUID:          613d592e-47f5-4206-96a7-210090d340ef Filesystem features:      has_journal ext_attr resize_inode dir_index filetype extent flex_bg sparse_super large_file huge_file uninit_bg dir_nlink extra_isize Filesystem flags:         signed_directory_hash  Filesystem state:         clean Filesystem OS type:       Linux Block count:              65536000 Reserved block count:     3276800 Free blocks:              64459851 First block:              0 Block size:               4096   65536000 total blocks * 4096 bytes/block = 268435456000 bytes = 268.4 GB = 250 GiB. It matches gparted.  3276800 reserved blocks = 13421772800 bytes = 13.4 GB = 12.5 GiB. It (again, sort of) matches Nautilus.  64459851 free blocks = 264027549696 bytes = 264.0 GB = 245.9 GiB. Why? Shouldnt it be either 250-12.5 = 237.5 (or 250-(12.5+4.11)=~233) ?  Removing reserved blocks:  $ sudo tune2fs -m 0 /dev/sda5 tune2fs 1.41.12 (17-May-2010) Setting reserved blocks percentage to 0% (0 blocks)  $ sudo tune2fs -l /dev/sda5 tune2fs 1.41.12 (17-May-2010) Filesystem volume name:   BACKUP Filesystem UUID:          613d592e-47f5-4206-96a7-210090d340ef Filesystem features:      has_journal ext_attr resize_inode dir_index filetype extent flex_bg sparse_super large_file huge_file uninit_bg dir_nlink extra_isize Filesystem flags:         signed_directory_hash  Filesystem state:         clean Filesystem OS type:       Linux Block count:              65536000 Reserved block count:     0 Free blocks:              64459851 Block size:               4096   As expected, same block count, 0 reserved blocks, but... same free blocks? Didnt I just freed 12.5 GiB ?  $ df -h /dev/sda5 Filesystem            Size  Used Avail Use% Mounted on /dev/sda5             247G  188M  246G   1% /media/BACKUP     Looks like I did. Avaliable space went up from 233 to 245.9 GiB. gparted didnt care at all, showing exactly same info! (useless to post an identical screenshot)  What a huge mess!  I tried to document it as best as I could... So, please can someone give me any clue on what's going on here?   What are those misterious 4.11 GiB missing from NTFS -> ext4 formatting? Why there are so many discrepancies between gparted, Nautilus, tune2fs, df? What is wrong with my math? (questions in bold scattered this post)   Any help is appreciated. While I can not figure what is going on, I am serilously considering giving up on ext4 in favor of NTFS for everything but my / partition.  Thanks! I am trying to find all the Sylow 2 subgroups of S4 using Sylow’s theorems. Now, I know that a Sylow 2 subgroup of S4 has size 8, and that there are either 1 or 3 of them (as the number of of Sylow 2-subgroups has form 1+2k and divides 3, the index).  Now my lecturer states “stabilisers of the 3 different bisections of {1,2,3,4} yield 3 distinct Sylow 2 subgroups”  Now my questions are:  1) What does it mean by stabilisers of 3 different bijections of {1,2,3,4}? Is that to say the elements of S4 which send {1,2} and {3,4} to {1,2},{3,4} i.e. by (12)(34)?  2) How does he know that there are 8 elements of this set? Perhaps this will become clearer after the 1st question is answered.  Many thanks, group theory is hard. What is the first movie to show this hard work message at the very end? What is the first movie to show this hard work message at the very end? How can you construct the bijection between real numbers and functions over naturals? Will tcpdump see packets that are being dropped by iptables? This is a question that arose in the comments here:   I normally hide footnotes from the TOC by doing this:  \section[title]{title\footnote{This is a footnote.}}.   This works in a lot of document classes (memoir, article, ...), but not in all of them. amsbook, for example, uses the optional argument only for the running heads so the footnote mark still ends up in the TOC. So what is the 'universal solution'? When will my device get the Android 4.2 update (Jelly Bean)? Mistakenly filed as Exempt from income tax witholdings for entire year 2019 I am a basic 30 year old filing Single, W2, healthy working individual. No dependents, no additional schedules, capital gains or losses other than Student Loan interest expense.   I started a new job in January 2019 and somehow during my new hire enrollment, "I or someone" marked in my file as Exempt... I am pretty familiar with filing my own 1040 (as its pretty basic)and I know for a fact that I would never for under any reason think that I am Exempt from absolutely anything tax related. Nobody in my new job's HR department reached out to confirm whether this Exemption was correct or not with me personally. My CPA said that anytime they receive an Exempt W-4 form, that they automatically return the form to the employer to confirm.   I unknowingly went 11 full months in 2019 without any taxes being withheld. Of course my paycheck was more than before, but I also got a better/higher salary job, so I never thought about the difference being from not paying taxes. And, no I do not print out my paystubs. It’s all through Paycom and it clearly shows you your gross pay &amp; a "$XXX" amount in deductions/withholdings - kicker is, you have to click the drop down menu to actually see the entire breakdown. At which point, I never saw the big $0 in taxes being withheld... I know the taxes have to be paid one way or another, but I don't feel like I should be entirely responsible for a $5k-$6k tax liability for a HR clerical error. Even if I did actually input it incorrectly the first time.   Is there a way to check with the IRS to see if they indeed receive my file as Exempt and if they did actually return it to my Employer? I’m just wondering if either HR forgot to follow up with me or even IF the IRS does actually cull these exempt requests in the first place? I don't have the liquidity to simply pay the IRS all at once.   What would be my best option going forward?