#coefficient — Public Fediverse posts

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

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https://www.europesays.com/nl/206659/ dit gaat Feyenoord verdienen in de Champions League #BreakingNews #BreakingNews #ChampionsLeague #coëfficiënt #DEKUIP #Dutch #eredivisie #FeaturedNews #FeaturedNews #Feyenoord #Headlines #Hoofdpunten #inkomsten #KnockOutfase #LaatsteVoetbalnieuws #LatestNews #LatestNews #miljoenen #Nederland #Nederlanden #Nederlands #Netherlands #News #Nieuws #NL #ticketverkoop #TopStories #TopStories #TvGelden #UEFA #voetbal #voetbalnieuws #Voorpaginanieuws

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#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Try our new word guessing game @ https://24hippos.com

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

A Tiny Forest of Resistors Makes for Quick and Dirty Adaptive Optics - The term “adaptive optics” sounds like something that should be really complicated... - https://hackaday.com/2022/07/25/a-tiny-forest-of-resistors-makes-for-quick-and-dirty-adaptive-optics/ #adaptiveoptics #interferometer #coefficient #mischacks #expansion #thermal #mirror #optics

One day, one decomposition
A121943: Numbers k such that the central binomial coefficient C(2k,k) is divisible by k^2

3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/A121943.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/A121943.html

#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #central #binomial #coefficient #divisible #graph #threejs #webGL

In the process I also thought of a #binomial #coefficient interpretation for choosing k things out of a bag of n objects when the objects can be put back and picked again. In probability problems, this is referred to as picking "with replacement".

Usually, when we say, n choose k, we do not allow repeated choices, every chosen object has to be chosen once. In this case, I want to allow replacements but at the same time, I want to keep using my trusted n choose k idea.

Here's my way around this: We'll still be picking k things, but we'll pick them not from 1 to n but from the set {1, 2, ..., 𝑛, 𝑟₁, 𝑟₂, ..., 𝑟ₖ₋₁}. That is, in addition to the n objects, we add what I'm calling "replacement tokens" 𝑟ₓ. If any of the 𝑟ₓ gets picked, then it is interpreted as the 𝑥th choice was put back and you are now choosing to pick that again. Since the 𝑘th choice is not put back, we only need replacement tokens for choices 1 to k-1.

With these replacement tokens, the problem becomes a standard choose k things out of this set, which we can resolve using the binomial coefficient to get: \({ n+k-1 \choose k }\).

I believe the standard approach to this is via #StarsAndBars but I liked the idea of this "replacement token". Admittedly, I didn't want to use stars and bars here and made up some stuff which I happened to like. :)

#math #counting #proof

#Activités #Fractions #cycle4

Choses promises, choses dues !
Je vous partage mes activités d'introduction sur les fractions, niveau #5e. 👨‍🏫

https://acloud11.zaclys.com/index.php/s/CLL3NWPzAFjC3LG

Je les ai toutes déjà menées, elles se sont toutes très bien passées. 👌
Reste la dernière à faire lundi qui est plus une introduction à la résolution de problème avec fraction (notamment avec des #proportions et donc de la #proportionnalité). 🤓
Ou comment faire de la proportionnalité sans #coefficient de proportionnalité. 🙃

📚🔍 #Rattrapages du #Bac en vue ? Pas de stress ! 😉
Voici quelques #conseils : ⤵️

Pour les rattrapages, visez les #matières où vous avez **honnêtement** eu le plus gros échec. 📉
Par exemple, si vous avez eu 12 en maths mais d'habitude vous avez 15, foncez sur les maths ! 🤓

Par contre, si vous avez eu 8 en philo et que votre moyenne est de 9, ça ne vaut peut-être pas le coup... 📖

Concentrez-vous sur les matières à fort #coefficient et maximisez vos #points.
Chaque point compte pour atteindre la moyenne ! 💪

Bon courage à tous ! 🌟

Du hast zwei oder mehr Variablen vor dir - und jetzt? 🤔 Um herauszufinden wie diese miteinander zusammenhängen, ist die Betrachtung der Korrelation bzw. des Korrelationskoeffizienten sinnvoll.
❗Aber Vorsicht: Korrelation bedeutet nicht unbedingt Kausalität! ➡ Es kann nur ein Zusammenhang festgestellt werden, keine Richtung!

#data4good #dataforgood #data4goodmonday #data #datascience #statistics #Statistik #correlaid #correlation #relationship #coefficient #positive #negative #analysis #ehrenamt

Je viens de le rendre compte que si je fais un #tableau de #proportionnalité pour une #vitesse donnée avec le temps en première lignes et la distance en seconde ligne.
Je perds la "formule" (v=d/t) mais le coefficient de proportionnalité pour passer de L1 à L2 est bien la vitesse.

À l'inverse, si je mets la distance en première ligne et le temps en deuxième ligne, je garde la "formule" (v=d/t) mais le #coefficient de proportionnalité pour passer de L1 à L2 est l'#inverse de la vitesse...

...😑

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

We would love to interview you! @ https://wordofthehour.org/interview

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------

Learn more languages @ https://wordofthehour.org/r/more

#coefficient : acting together to produce an effect

- French: coefficient

- German: der Beiwert

- Italian: coefficente

- Portuguese: coeficiente

- Spanish: coeficiente

------------------

Thank you so much for being a member of our community!

One day, one decomposition
A121943: Numbers k such that the central binomial coefficient C(2k,k) is divisible by k^2

3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/A121943.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/A121943.html

#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #central #binomial #coefficient #divisible #graph #threejs #webGL

One day, one decomposition
A121943: Numbers k such that the central binomial coefficient C(2k,k) is divisible by k^2

3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/A121943.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/A121943.html

#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #central #binomial #coefficient #divisible #graph #threejs #webGL

One day, one decomposition
A121943: Numbers k such that the central binomial coefficient C(2k,k) is divisible by k^2

3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/A121943.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/A121943.html

#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #central #binomial #coefficient #divisible #graph #threejs #webGL

One day, one decomposition
A121943: Numbers k such that the central binomial coefficient C(2k,k) is divisible by k^2

3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/A121943.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/A121943.html

#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #central #binomial #coefficient #divisible #graph #threejs #webGL

In the process I also thought of a #binomial #coefficient interpretation for choosing k things out of a bag of n objects when the objects can be put back and picked again. In probability problems, this is referred to as picking "with replacement".

Usually, when we say, n choose k, we do not allow repeated choices, every chosen object has to be chosen once. In this case, I want to allow replacements but at the same time, I want to keep using my trusted n choose k idea.

Here's my way around this: We'll still be picking k things, but we'll pick them not from 1 to n but from the set {1, 2, ..., 𝑛, 𝑟₁, 𝑟₂, ..., 𝑟ₖ₋₁}. That is, in addition to the n objects, we add what I'm calling "replacement tokens" 𝑟ₓ. If any of the 𝑟ₓ gets picked, then it is interpreted as the 𝑥th choice was put back and you are now choosing to pick that again. Since the 𝑘th choice is not put back, we only need replacement tokens for choices 1 to k-1.

With these replacement tokens, the problem becomes a standard choose k things out of this set, which we can resolve using the binomial coefficient to get: \({ n+k-1 \choose k }\).

I believe the standard approach to this is via #StarsAndBars but I liked the idea of this "replacement token". Admittedly, I didn't want to use stars and bars here and made up some stuff which I happened to like. :)

#math #counting #proof