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: https://www.whatdotheyknow.com/request/queue_allocation_for_britishnon.

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 (https://www.gov.uk/topic/immigration-operational-guidance) 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.

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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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