am carrying out research to look at differences in fish density and fish species richness when using two different underwater visual census methods. My data was originally count data but then converted to fish density and so decided to use an offset using the following formula:   model1 &lt;- glm(g_den ~ method + site + depth + offset(log(area)), poisson)  My 3 predictor variables are method, site and depth which I ordered as factors when I input them.  My response variables are grouper species richness, grouper density and the same for other fish groups.   Despite using the offset i keep getting the same error, could anyone advise me why this is?  In dpois(y, mu, log = TRUE) : non-integer x = 0.037500
I am carrying out research to look at differences in fish density and fish species richness when using two different underwater visual census methods. My data was originally count data but then typically this is changed to fish density but I have still decided to use a Poisson GLM, which I hope is right.   model1 &lt;- glm(g_den ~ method + site + depth, poisson)   My 3 predictor variables are method, site and depth which I ordered as factors when I input them.   My response variables are grouper species richness, grouper density and the same for other fish groups. I am aware that density is not an integer and it is numerical data e.g 1.34849. I am now however getting this error:      In dpois(y, mu, log = TRUE) : non-integer x = 0.037500   I've been reading up and many people suggest using an offset, is this the most advisable thing to do?
How to stop Katoolin from crashing Ubuntu? I updated from 14.04 to 16.04 and used katoolin to add some Kali tools. After restart, the boot up of Linux ran codes, but stopped at a certain point never to load login. So I reinstalled 16.04, ran updates and tried again. This time, the boot menu was changed to Kali as well as the OS name (I'm using Windows 7 dual boot with Ubuntu); login started in Ubuntu style without user name, had to type in the name... then it goes to the logo screen never to log in.  How can I use katoolin without it affecting the boot menu interface? Does this happen when the user selects the option 'add Kali repositories and updates'? Are the Kali repositories necessary? I'd much rather just install the tools and not mess with the Ubuntu interface. OneDrive isn't uploading all my photos from my phone I've got an Office 365 (Microsoft 365) Family license. I've set up OneDrive on my Samsung Galaxy S10 phone. I've verified that it is set up by following the . I've got it set up to upload photos using either Wi-Fi or mobile. And I see photos I've taken with my phone, uploaded to OneDrive from as recently as a few days ago. So, it works. But not all the time or for all my photos. I've scrolled through my photos on my phone's Gallery app. I see some photos from as recently as 4 days ago to several months ago, are not uploaded to OneDrive. Why is that? Triggered send status queued, need to get an alert If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative. If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.  I need to proof this theorem and I have something like this below. However, I do not know how to continue this proof.   $(x+y)^{2}-(x+y)=x^{2}+xy+yx+y^{2}-x-y=(x^{2}-x)+(y^{2}-y)+xy+yx\in Z(R)$  so $xy+yx\in Z(R)$ Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition. Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition. If you finish a world without fighting the Broodal, can you still fight them later? I've just seen that in a speedrun: If you collect enough moons to move on to the next world, and cash them in at the ship, the broodal ship just leaves.  Does that mean that the Broodal fight wont happen, so you missed the triple moon it gives? Or if you go to the arena, even without the ship you will have the fight happen?  Clip where this happens : Is the Formula Logically Valid? I have a question for my exam and I find it hard to understand.  I have to prove that the following formula is logically valid:    The professor told me to "push" all the symbols inside the brackets, and use the deduction theorem.  But I don't know how to do it, because I can't find the identities to push the "exist" symbol inside the brackets.  Your help is appriciated, thank you.  Alan Countable sets have the same cardinality as the natural numbers (or a subset of the natural numbers, depending on who you ask).  Can one make the same claim about uncountable sets and real numbers?  In other words, do all uncountable sets have same cardinality as real numbers? Indefinite article and people's names Sometimes, on the internet, particularly in online games, I see people using the  indefinite article before someone's name: "I see a Joey" or "I hug a Polly". I know some of these people and I'm positive that they are from English-speaking countries, which means it is not a mistake, they do it intentionally.  I was always taught that articles are not used in front of people's names. Maybe it is some sort of slang? What does it mean, when the indefinite article is put in front of someone's name?  Edit: There is a group of people standing somewhere. And somebody, named Joey, appears. They all know Joey. And someone in this group says "I see a Joey". So it is used when someone sees his or her friend. But I don't understand why they use the indefinite article here. I was listening to the song  by Eminem. I've listened it before many times but from the time I've started learning English grammar I am getting confused while reading or listening English. This time I got stuck at the line "Too much pride, between you and I Not a jealous man but females lie". The phrase "between you and I" always sounded correct to me. But recently I learned that me is only used as an object and hearing that phrase sparked my head. Then I searched over the net and found on  that it is actually a hypercorrection.  says,      "...according to many grammarians and stylists a pronoun in a prepositional phrase in English is supposed to be in the oblique case...   ...though there is still disagreement on whether the phrase itself in today's language is grammatically correct or not."   My first question is, why is there a disagreement? Don't we always use the object form in the oblique case? If not then what are those cases?  Secondly, I want to know how common the phrase "between you an I" is? Is it used colloquially by common native English speakers. By common speakers I mean who don't bother grammar. I searched on  and got some results, e.g. . I found the phrase in another song -- . Is it a common mistake among native English speakers or is it used to give some special linguistic affect? I think this phrase has some linguistic affect that is why the singer would have given the name of her song "Between You and I". Because no one would like to use incorrect English in the title of a song, it must have some purpose. Word for a temporary project name I want to build a personal portfolio so my questions related are as follows:   What should I include in the portfolio apart from Home, About, Work, Contact? I have never worked on any real life examples so what to display in the work/portfolio page? What kind of design samples do I need to include? Ex: Website, Logo, Banner etc   Any help is appreciated.  Thanks! Prove the inequality $n! \geq 2^n$ by induction I'm having difficulty solving an exercise in my course.  The question is:      Prove that $n!\geq 2^n$.   We have to do this by induction. I started like this:   The lowest natural number where the assumption is correct is $4$ as: $4! \geq 2^4 \iff 24 \ge 16$. The assumption is: $n! \ge 2^n$.   Now proof for $(n+1)$ which brings me to: $(n+1)! \ge 2^{(n+1)}$  I think I can rewrite it somehow like this:  $$ {(n+1)} \times {n!}  \stackrel{\text{(definition of factorial)}}{\ge} 2^n \times 2 $$  $$ (n+1) \times 2^n \ge 2^n \times 2 $$  Then I think I can eliminate the $2^n$ and have something like this: $n+1 \ge 2$, or $n \ge 1$.  But I think I'm wrong here some where and was hoping somebody has some advice on this. How can I prove the above assumption?  Any help would be appreciated, kind regards. I've been given a proof that shows the following:  If $f:[a,b]\to \mathbb C$ is a continuous function and $f(t)=u(t)+iv(t)$ then $$\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$$ The proof begins by letting $\theta$ be the principle argument of the complex number $\int_a^b f(t)dt$ and there is one step in the proof I don't understand; can anyone explain to me why $$\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)$$  Would this not imply that the imaginary part of the left hand side is equal to $0$? I'm not sure why the LHS and RHS would be equal here. Best compiler warning level for C/C++ compilers? "invalid use of incomplete type" error with partial template specialization Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &amp;= \sqrt {x^2} \\ f(-1) &amp;= \sqrt {(-1)^2} \\ f(-1) &amp;= \sqrt {1} \\ f(-1) &amp;= 1 \end{align}$  $\begin{align} g(x) &amp;= x \\ g(-1) &amp;= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$? Differentiation - simple case In the book calculus made easy, page 22 the case of the negative power for   $y=x^{-2}$    $$\begin{align}  y+dy &amp; =(x+dx)^{-2}\tag{1}\\ \\ &amp; = x^{-2}\left(1+\frac{dx}{x}\right)^{-2}\tag{2} \end{align}$$  How do they get from $(1)$ to $(2)$?  And why isn't the binomial function applied to the second line but applied on the third line (the line that comes after line $(2))$? The top bar is showing (again) the suggested edits count and 10K flags count, like it was before.  On Stack Overflow it looks like this: (that's the original color, same as in old top bar)        And on this very Meta site:        Same top bar. But different colors! Consistency broken.  I've just noticed the suggested edits count on Meta:        Black on black, I can see just a floating number - really confusing and out of any context.  In the name of Consistency, can we please have the same color on all sites? nginx open reverse proxy? Somehow, and I'm not sure how, we have configured an nginx reverse proxy that is acting as an open proxy.  The log files tells us the matching requests are going against the following conf file:  map $http_upgrade $connection_upgrade {         default upgrade;         ''      close; }  access_log      /var/log/nginx/access.p.log; error_log       /var/log/nginx/error.p.log;  proxy_next_upstream     error timeout invalid_header http_500 http_502 http_503 http_504; proxy_buffers           8 32k; proxy_buffer_size       64k; proxy_http_version      1.1; proxy_set_header        Host            $host; proxy_set_header        X-Real-IP       $remote_addr; proxy_set_header        X-Forwarded-For $proxy_add_x_forwarded_for; proxy_redirect          off; proxy_set_header        Upgrade         $http_upgrade; proxy_set_header        Connection      $connection_upgrade;  server {         listen          80;         server_name     ~first.*\.second\.com;          location / {                 resolver                172.16.1.1 172.16.1.2;                 proxy_pass              http://$host;                 proxy_read_timeout      7200s;         } }  server {         listen          443;         server_name     ~first.*\.second\.com;         ssl             on;          location / {                 resolver                172.16.1.1 172.16.1.2;                 proxy_pass              https://$host;                 proxy_read_timeout      7200s;         } }   The top-level conf file looks like this:  user nginx; worker_processes 8; pid /var/run/nginx.pid;  events {         worker_connections 4096;         # multi_accept on; }  http {      ##     # Basic Settings     ##      sendfile on;     tcp_nopush on;     tcp_nodelay on;     keepalive_timeout 65;     types_hash_max_size 2048;     # server_tokens off;      # server_names_hash_bucket_size 64;     # server_name_in_redirect off;      include /etc/nginx/mime.types;     default_type application/octet-stream;      ##     # Logging Settings     ##      access_log /var/log/nginx/access.log;     error_log /var/log/nginx/error.log;      ##     # Gzip Settings     ##      gzip on;     gzip_disable "msie6";      # gzip_vary on;     # gzip_proxied any;     # gzip_comp_level 6;     # gzip_buffers 16 8k;     # gzip_http_version 1.1;     # gzip_types text/plain text/css application/json application/x-javascript text/xml application/xml application/xml+rss text/javascript;      ##     # nginx-naxsi config     ##     # Uncomment it if you installed nginx-naxsi     ##      #include /etc/nginx/naxsi_core.rules;      ##     # nginx-passenger config     ##     # Uncomment it if you installed nginx-passenger     ##      #passenger_root /usr;     #passenger_ruby /usr/bin/ruby;      ##      include /etc/nginx/conf.d/*.conf;     #include /etc/nginx/sites-enabled/*;   }  what here is causing this?