Customs and immigration – Can a non-EEA passport holder with a biometric residence permit (BRP), who accompanies the EEA family, join the EEA quota at the borders of the United Kingdom?

I believe this May 8, 2019, Freedom of Information response raises an "official" light on this issue:

It says that

".. Passengers who are not from the EU or the EEA and travel as family members of a British citizen or an EU / EEA national may join the EU / EEA queue. subject to local queue management arrangements, .. "

This suggests that users must independently identify these "local queue management agreements" from the specific border sites.

I would be grateful if someone ( could prove from the official documents or other official state sources that borders can actually set their own queuing on-site.

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statistics – Show that a Weibull distribution belongs to an exponential family

I am studying statistics and have encountered a problem where I have some problems that worry me.

I got the density function of a Weibull distribution

I shall show that the Weibull distributions with fixed k belong to the exponential family with the form:

fθ (y) = exp (a (y) b (θ) + c (θ) + d (y))

I have taken the logarithm of the function and have:

log f (y; λ, k) = log (k / y) + (k-1 * log (y / λ)) -k (y / λ))

I am not sure how to derive the canonical parameter and determine where the terms belong. All tips on arranging the logs are welcome.

Natural and safe alternative to bring health and well-being to your furry family members

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Irreducibility of a family of hypersurfaces via $ mathbb {F} _p $

To let $ Q in mathbb {F} _p (x, y, z) $ be a geometrically irreducible polynomial. If it helps, accept $ Q $ is a square shape. Note the parameterization

$$ x = u (r), y = v (s), z = w (t), $$
from where $ u, v, w $ are a variable polynomial over $ mathbb {F} _p $, Under what conditions (hopefully mostly dependent on $ Q $) we can say that $ Q $ remains geometrically irreducible. All I need is that there is a one-dimensional irreducible component under the Frobenius map.

There is a red herring: take $ Q (x, y, z) = y ^ 2 – xz $
$$ x = r ^ 2, z = at ^ 2, a not in mathbb {F} _p ^ 2. $$

Then $ Q $ is geometrically irreducible, but after paratisation we get
$$ y ^ 2 – a (rt) ^ 2 = (y – sqrt {a} rt) (y + sqrt {a} rt). $$
I feel that this is a special situation.

What would you say to the Koch family and the body of David Koch if you could speak at the satanic funeral of David Koch?

I would say, "You know the flames that you felt when you sadly escaped death in this plane crash in 1991? Now enjoy this feeling for eternity. Charles Koch – it's time for you and your family to join your brother. "

When I speak of the dead, the dead person does not automatically become a saint. Saying bad things about Hitler did not make him a saint. Koch is not far from Hitler

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Not so long ago, searching astronauts discovered a friendly planet with very cute creatures in the universe. These creatures called themselves the Zuks and the planet Zukia. They proved to be very smart and hardworking, and the planet is very rich in rare and expensive minerals in the universe. Thus began a mutually beneficial partnership between the Zuks and other representatives of the universe.