In the movie The Avengers (2012), at the end the Hulk smashes Loki to the floor. After that we can see Loki injured with blood on his face.     How can the Hulk hurt a god? Shouldn't Loki be super strong, how can a god bleed?
In Thor: The Dark World, Odin and Loki are discussing Loki's actions during The Avengers.  Loki argues that he would have ruled Earth as &quot;a benevolent god&quot;, and Odin replies:  Odin: We are not gods! We are born, we live, we die, just as humans do.  Now that's pretty definitive.  And Loki's snarky reply (&quot;Give or take five thousand years...&quot;) would seem to confirm that Asgardians have a limited lifespan, and will eventually die of old age.1  And we know that Asgardians can die in battle, because we've seen it happen over and over with supporting characters and Asgardian soldiers. Now, I've always been more of a fan of Thor stories (like Jason Aaron's Thor: God of Thunder arc) in which Thor and the Asgardians are gods.  Straight-up, genuine, higher-level-of-existence gods, not von-Däniken-esque wannabes playing on the tired old notion of magic just being super-advanced science.  BUT, despite my preferences, this exchange would seem to confirm once and for all that in the MCU, the Asgardians are not gods, but rather were mistaken for gods because of their ram-horns full of awesome-sauce and their magnificent bulging pecks. But wait!  Watch that movie for about 30 seconds longer and we come across this exchange:  Thor: She is ill. Odin: She is mortal. Illness is their defining trait.  Wait, what?  I thought Asgardians weren't immortal?  They can die in battle, and they naturally die of old age.  In my book, that makes them long-lived, but entirely &quot;mortal&quot;.  Surely Odin isn't splitting a hair so fine as to say that resistance to disease alone makes one &quot;immortal&quot;? So, if the Asgardians of the MCU aren't actually gods (grumblegrumblegrumble...), and they can't live forever (or even particularly long in a cosmic sense), and they can be killed in battle, in what sense are they &quot;immortal&quot; at all?  1 Fun fact: adjusted for a roughly 5,000 year lifespan, 1,048-year-old Loki is currently in the tail-end of his rebellious teenage phase.
Why does flag marking as helpful/declined not always correlate with moderator action? I'm having trouble understanding what I'm doing wrong here: I'm trying to set up two network Locations, one for my home network via my cable modem and a second for Wi-Fi tethering via my mobile phone. What I'd like to be able to do is change the Location (under the Apple menu &gt; Location) and have my Mac automatically connect to the proper Wi-Fi network: either my home network SSID or my WiFi tethering SSID. I defined two locations, switched between them, and set the preferred network list for each location. But when I change the preferred network list for one Location, it changes the list for the other Location as well! Am I doing something wrong or is it really not possible to maintain a separate list of Preferred Networks for each network Location? It seems doubtful it's not possible since just about every other netrok setting depends on the location... I am using OS X 10.7.2. For the Fibonacci numbers, show for all $n$: $F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$ The definition of a Fibonacci number is as follows:  $$F_0=0\\F_1=1\\F_n=F_{n-1}+F_{n-2}\text{ for } n\geq 2$$   Prove the given property of the Fibonacci numbers for all n greater than or equal to 1. $$F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$$  I am pretty sure I should use weak induction to solve this. My professor got me used to solving it in the following format, which I would like to use because it help me map everything out...  This is what I have so far:  Base Case: Solve for $F_0$ and $F_1$ for the following function: $F_nF_{n+1}$.  Inductive Hypothesis:  What I need to show: I need to show $F_{n+1}F_{n+1+1}$ will satisfy the given property. Proof Proper:  (didn't get to it yet)  Any intro. tips and pointers? I find that the minimum in-call volume on my HTC Desire is still far to loud on some calls.  Since I got the phone, I have flashed several different custom ROMs, radio updates, etc, and have had the issue throughout.  Do you know of any way (setting, hack, app, whatever) to reduce the minimum volume?  My device is rooted and I am currently running one of the CyanogenMod 7 (Gingerbread) nightly builds. In Catan, can you continue a "circular" road? If I have 6 roads that go all around a single hex tile, they are a continuous uninterrupted road (assuming none of my opponents interrupts it with their settlements/cities). But, can I continue this 6-piece road and create a 7-piece continuous road by building a road that goes out of this hex tile? Here's an image of what I'm talking about:    Is this a valid 7-piece continuous road? Prove Uncountable set minus a countable set is uncountable Let B be a countable set, and let A be an uncountable set. Prove that A-B is uncountable How do you become friends with someone? Let $\phi : R \to S$ be a homomorphism. If $N$ is an ideal of $S$, then $\phi ^{-1} (N)$ is an ideal of $R$. How to Update Multiple Array Elements in mongodb In how many ways can a number be expressed as a sum of consecutive numbers? How can one differentiate between Dr. (PhD) and Dr. (MD or DO)? I have a simple question regarding "conditional probability" and "Likelihood". (I have already surveyed this question  but to no avail.)  It starts from the Wikipedia . They say this:     The likelihood of a set of parameter values, $\theta$, given   outcomes $x$, is equal to the probability of those observed outcomes   given those parameter values, that is      $$\mathcal{L}(\theta \mid x) = P(x \mid \theta)$$   Great! So in English, I read this as: "The likelihood of parameters equaling theta, given data X = x, (the left-hand-side), is equal to the probability of the data X being equal to x, given that the parameters are equal to theta". (Bold is mine for emphasis).  However, no less than 3 lines later on the same page, the Wikipedia entry then goes on to say:     Let $X$ be a random variable with a discrete probability distribution   $p$ depending on a parameter $\theta$. Then the function      $$\mathcal{L}(\theta \mid x) = p_\theta (x) = P_\theta (X=x), \, $$      considered as a function of $\theta$, is called the likelihood   function (of $\theta$, given the outcome $x$ of the random variable   $X$). Sometimes the probability of the value $x$ of $X$ for the   parameter value $\theta$ is written as $P(X=x\mid\theta)$; often   written as $P(X=x;\theta)$ to emphasize that this differs from   $\mathcal{L}(\theta \mid x) $ which is not a conditional probability,   because $\theta$ is a parameter and not a random variable.   (Bold is mine for emphasis). So, in the first quote, we are literally told about a conditional probability of $P(x\mid\theta)$, but immediately afterwards, we are told that this is actually NOT a conditional probability, and should be in fact written as $P(X = x; \theta)$?   So, which one is is? Does the likelihood actually connote a conditional probability ala the first quote? Or does it connote a simple probability ala the second quote?  EDIT:  Based on all the helpful and insightful answers I have received thus far, I have summarized my question - and my understanding thus far as so:    In English, we say that: "The likelihood is a function of parameters, GIVEN the observed data." In math, we write it as: $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x)$. The likelihood is not a probability.  The likelihood is not a probability distribution.  The likelihood is not a probability mass. The likelihood is however, in English: "A product of probability distributions, (continuous case), or a product of probability masses, (discrete case), at where $\mathbf{X} = x$, and parameterized by $\mathbf{\Theta}= \theta$." In math, we then write it as such: $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = f(\mathbf{X}=x ; \mathbf{\Theta}= \theta) $ (continuous case, where $f$ is a PDF), and as $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = P(\mathbf{X}=x ; \mathbf{\Theta}= \theta) $ (discrete case, where $P$ is a probability mass). The takeaway here is that at no point here whatsoever is a conditional probability coming into play at all.  In Bayes theorem, we have: $P(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = \frac{P(\mathbf{X}=x \mid \mathbf{\Theta}= \theta) \ P(\mathbf{\Theta}= \theta)}{P(\mathbf{X}=x)}$. Colloquially, we are told that "$P(\mathbf{X}=x \mid  \mathbf{\Theta}= \theta)$ is a likelihood", however, this is not true, since $\mathbf{\Theta}$ might be an actual random variable. Therefore, what we can correctly say however, is that this term $P(\mathbf{X}=x \mid \mathbf{\Theta}= \theta)$ is simply "similar" to a likelihood. (?) [On this I am not sure.]   EDIT II:  Based on @amoebas answer, I have drawn his last comment. I think it's quite elucidating, and I think it clears up the main contention I was having. (Comments on the image).    EDIT III:   I extended @amoebas comments to the Bayesian case just now as well: The units of $\mathbb Z[\sqrt{2}]$ How can I show that the units $u$ of $R=\mathbb Z[\sqrt{2}]$ with $u&gt;1$ are $(1+ \sqrt{2})^{n}$ ?   I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+\sqrt{2}$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help. Two very closely related questions:  If a card has two different "Sacrifice this creature for X" abilities on it, does sacrificing it give both benefits?  Likewise, if a card requires a creature to be sacrificed, but that creature has a "Sacrifice this creature for X" ability, do you get the sacrifice benefit when you sacrifice it for the other requirement? Since drush 9 (drupal 8.4 requirement), alias @sites does not seem to be implemented anymore (or yet ?).  $ drush @sites cr The alias @sites could not be found.   Is this a way to get the same behavior that does not consist of typing my 300 alias in a config file ? How To Change The Fonts In Windows 8 In Windows 7, there is an option to change the fonts for things such as the active title bar, menu, etc. It was found by opening the Window Color and Appearance settings in the Control Panel. I just installed Windows 8 and would like to change the fonts, but I can't find anything to change the actual fonts (there is however a limited drop-down list to change a few font sizes in the appearance settings). Does anyone how to change the fonts used in Windows 8? How should I phrase an important question that I need to ask a professor? When needing to ask a professor about something important and potentially delicate, what is an appropriate way to phrase the question or write the email?  Examples include:   Requesting a recommendation letter Asking for exceptions to policy (e.g., ignoring a course prerequisite, rescheduling an exam) Asking if you can do research with them, join a group, get into a program Scheduling a meeting, exam, or defense Letting them know about a change of plans Letting them know about a life event affecting your work or studies (e.g., wedding, funeral, sickness, vacation) Use fixed keyboard for a given app I would like to tell my Android that when I'm in a given app it should switch to a given keyboard. My use case is that I want to use Hacker keyboard with Connectbot. Symmetric Inequality in $\mathbb{R}$ Sometimes, others just inform us about things without us asking them, other times they do not do that even if we ask them. I want a verb for when others do not give us information — whether because they do not like to or they just don't give it, but we cannot directly ask them to — but we actively extract the information from them sometimes without them being aware of it.   So the word I am looking for shouldn't be only for cases where we coerce or force others to give us information, and it shouldn't be only for cases where others do not like to give us information.  Example:     I needed that information. But I couldn't ask her for it, or most   probably she would not give to me even if I asked her. So, I ....   the information I wanted.   It could be a phrasal verb and so not just a single word, but I couldn't devise a  blank suitable for all possibilities!   Two verbs I've already had but I think are not suitable:  to give someone the third degree   To prize something from/out of somebody