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Closing Questions As Not Clear Without Giving Feedback
I believe this is 100% in keeping with the purpose of putting questions on hold, to "pause" them until they can be answered properly. Jumping in with even more comments (especially if they would just be reiterating the current ones) doesn't seem either appropriate or necessary to me
• Related Questions
Is removing network access to a shared Time Machine drive mid-backup similar to unplugging a physical drive?After trying WiFi based Time Machine backups on a NAS (MyCloud) and a NAS-like arrangement (USB hard drive plugged into Airport Extreme), I have found no solution where Time Machine 'just works' with disconnected laptops. Time Machine works perfectly for desktops, Ethernet attached or otherwise. But when used on a laptop, where the user can interrupt Time Machine, I have found hung, stalled, and corrupt Time Machine images frequently. Typically, Time Machine will report a problem with the backup, and start a new one.The only solution I found was to stop using Time Machine, and instead use a cloud-based backup system that was built assuming this use case was normal. I prefer Crashplan, but there are many others such as Backblaze, etc. My assumption is that Time Machine was built assuming always connected hard drives, and Wifi is not a well supported case. Time Capsule perhaps has some secret sauce, but I have no experience with it
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Closing a question should force leaving a note duplicate
The closing reasons are reported in the FAQ, and the description is quite clear.The same description appears when a question is closed, and the OP is able to see the reason why the question has been voted to be closed even before the necessary number of users vote for closing it.If a user keeps asking questions that are closed, then that user should try to understand why the questions have been closed; asking on the meta site is a start.Forcing who is voting to close to leave a comment about the reason they are voting to close doesn't seem necessary, and it could just cause a flow of comments about the question not deserving to be closed, or deserving to be closed.
In some cases it's not necessary to further explain why the question is being voted to be closed. What would a comment about the question being closed as duplicate of another one add?
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Small ideas that became big
Pick's theorem states that the area $A$ of a polygon $P$ in $mathbbR^2$, whose vertices are in the lattice $mathbbZ^2$, can be computed by means of the formula
$$AIfracB2-1,$$
where $I$ is the number of lattice points in the interior of $P$, and $B$ is the number of points in the boundary of the polygon $P$. George Pick published this theorem in 1899, in his article "Geometrisches zur Zahlenlehre" Sitzungsberichte des deutschen naturwissenschaftlich-medicinischen Vereines fr Bhmen "Lotos" in Prag. (Neue Folge). 19: 311319. Apparently, this result remained unknown until the middle of the 20th century when Hugo Steinhaus included it in his book "Mathematical Snapshots".This beautiful result is a precursor of theories about "counting lattice points in polyhedra" (e.g., Ehrhart's theory, and generalized Euler-MacLaurin summation formulas) which intersect, as far as I know, with problems in linear programming, values of number theoretic zeta functions, toric varieties, and even physics (I've heard).
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How to avoid the need to set fonts every time in notepad
It sounds like the Notepad does not have permissions to write the configuration file. Try running it as an administrator: Either right click and select run as administrator, or hold down CtrlShift while launching the application. While it running as an administrator, make the changes and save and close the application.
This should not occur in general, you should have sufficient permissions to read/write to your home directory where the configuration files are written (usually under C:Users%user_name% ). I would recommend uninstalling the Notepad, and then reinstalling it. I believe in the installation there is option to change the location of the user settings, make sure it is in a location that your user has permissions to access. In case steps do not fix the problem, your user might be corrupt, in which case you could make a new user and see if this fixes the problem, though this should be a last resort
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How to derive ecliptic in simplified solar system, with vector math?
Let $vec w$ be the projection of $vec r$ onto the ecliptic plane: we know that the angle between $vec w$ and $vec r$ is $alpha90-23.4466.56$, hence $vec w$ lies on a cone of axis $vec r$ and half-width $alpha$.On the other hand, from the given data we can compute the angle $theta$ between $vec r$ and $vec s$:
$$thetaarccosleft(vec rcdotvec sover|vec r||vec s|right).$$
The angle $beta$ between $vec w$ and $vec s$ is then given by $$betaarccosleft(costhetaovercosalpharight),$$
hence $vec w$ lies on a cone of axis $vec s$ and half-width $beta$.But those two cones have two rays in common: one of them contains $vec w$ but the other doesn't. Without some additional info I'm afraid you cannot tell which ray is the good one, that is you have in general two possible choices for the ecliptic plane.
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Bari BRI to Rome Fiumicino FCO Ryanair On Time Peformance - Is this doable?
Concerning your question about the camera bag: if it fits under the seat, then you can always argue with the gate agents that you will put it there. Usually they only mark large suitcases to be checked in instead of carry-on. Boarding early also reduces the probability of such a procedure. Instead please not that if your flight from BRI to FCO is delayed considerably you will not only be unable to pick up your luggage but might even miss your FCO to AMS flight. That is a risk I would try to avoid. Could you travel a day earlier instead? And as for you true question: if the flight is on time you have plenty of time to pick up your luggage but in case it is not you are in some trouble. How much would be the cost to check in your luggage to BRI and back compared to luggage storage at FCO in the end?.
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Construct a ring containing $16$ element where EVERY element $rneq 0$,$1$ is a zero divisor
In a finite commutative ring $R$, every noninvertible element is a zero divisor.Indeed, if $r$ is not invertible, the map $xmapsto rx$ is not surjective (because $1$ is missed), so it is not injective. Being a group homomorphism with respect to addition, its kernel is nonzero, which proves the claim.
Your ring must have a single invertible element, hence its characteristic needs to be $2$: indeed, $-1$ is invertible, so we need $-11$.Thus $R$ has to be a four dimensional vector space over $mathbbF_2mathbbZ/2mathbbZ$ and it's quite natural to look for a ring structure on $mathbbF_2^4$.One possibility is the product ring, which indeed satisfies the requirement, because an element of the form $a(a_1,a_2,a_3,a_4)$ where one of the entries is zero is surely a zero divisor: if $e_i$ denotes the $4$-tuple with $1$ at the $i$-th place an zero elsewhere, then $a_i0$ implies $ae_i0$.
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How did the timeline work?
Based on your question, we're ignoring the fun times around Seth at the start of the film, and focusing on Joe. Having seen the film twice, there are some nuances I missed the first time through, so I'm altering my answer. There are three timelines that matter:The director acknowledges there are some oddities around the timelines, as the Rainmaker is described as having a fake jaw and seeing his mother killed in front of him, which would suggest that we always see timelines where Old Joe shot widdle Rainmaker.
Within the film narrative, I think it's stating that one of the ramifications of timeline 1 above was that because Joe knew his loop would be closed, that he then was able to successfully overpower the henchmen and choose to go back in time. Thus, he created the 2nd timeline, because the causality loop of him killing himself wasn't tight enough, due to the possibility of overpowering the henchmen
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Does coordinate time have physical meaning?
Proper time represents the physical aging of a massive particle, and by this it is the only time which is to take into account for the physical description of a particle.But coordinate time is not without physical meaning: There would be no detection of events without coordinate time. When two particles are traveling through the same place in space, their proper time will not provide the information if it happened simultaneously, i.
e. that they encountered, i.
e. that there is an event. For this information you need the Minkowski diagram of at least one of both particles, and by the way the Minkowski diagram of any observer includes the coordinates of both particles, providing the information if they did encounter or if they did not.Minkowski diagrams are showing the coordinate time of all particles (with different simultaneities). In contrast, it is not possible to represent the proper time of two different frames in one diagram.
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Determining whether a coin is fair
The total number of successes in $n118$ runs is binomial $(n,frac12)$ hence the probability $p_n(k)$ to get at least $k84$ successes is
$$
p_n(k)2^-nsum_ik^nnchoose i.
$$
When $k$ is significantly larger than $fracn2$, $p_n(k)$ is very small and an estimation of how small $p_n(k)$ is is obtained through a large deviations estimate. This says that $p_n(k)leqslant p_n^*(k)$ with
$$
p^*_n(k)2^-ninf(1s)^ns^-k,;,sgeqslant1.
$$
For every $kgtfracn2$, the infimum is reached at $sfrackn-k$, hence
$$
p^*_n(k)2^-nn^nk^-k(n-k)^-(n-k)left(Ileft(tfracknright)right)^-n,quad I(t)2t^t(1-t)^1-t.
$$
For example, if $k84$ and $n118$, then $t.712$ hence $I(t)approx1.09710$ and
$$
p^*_118(84)approx(1.09710)^-118approx10^-5.
$$
Numerically, $p_118(84)approx2.36224cdot10^-6$ and $p^*_118(84)approx1.78153cdot10^-5$.
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How to calculate mean value on cycle space?
This is a topic referred to as the mean of circular quantities. First convert the data into angles, for example if you have a number of hours $0 leq h$$
theta frac2pi 24 h
$$
Then from a set of angles $theta_i$ we can compute the mean using
$$
bartheta atan2(sum_i sin theta_i, sum_i cos theta_i)
$$
which can be understood by imagining putting points on a circle at the given angles, finding the average position of these points (which will be inside the circle), and then finding the angle to this average point. This explains one of the issues you have in your question, if the 2 points you want to average are on opposite sides of the circle, the average point will be the center of the circle, which results in an undefined angle. Finally, convert back to the original units
$$
barh frac24 2pi bartheta
$$
