WordofTheHour#coefficient : acting together to produce an effect
- French: coefficient
- German: der Beiwert
- Italian: coefficente
- Portuguese: coeficiente
- Spanish: coeficiente
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#coefficient : acting together to produce an effect
- French: coefficient
- German: der Beiwert
- Italian: coefficente
- Portuguese: coeficiente
- Spanish: coeficiente
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Word of The Hour's Annual Survey @ https://wordofthehour.org/r/form
Word of The Hour - Annual Survey (2025)
Your responses to the questions below will directly impact the future of Word of The Hour. Your support and kindness has really meant a lot over the past three years. Thank you so much! Michael Wehar https://wordofthehour.org [email protected]
#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
Word of The Hour - Annual Survey (2025)
Your responses to the questions below will directly impact the future of Word of The Hour. Your support and kindness has really meant a lot over the past three years. Thank you so much! Michael Wehar https://wordofthehour.org [email protected]
#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
Word of The Hour - Annual Survey (2025)
Your responses to the questions below will directly impact the future of Word of The Hour. Your support and kindness has really meant a lot over the past three years. Thank you so much! Michael Wehar https://wordofthehour.org [email protected]
Rémi EismannOne 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
Cyril I. 🧮📐📚🔍 #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 ! 🌟
Dinah GalligoLe Monde #Bac2024 : « Le faible #coefficient affecté à l'épreuve de #philosophie est une anomalie » https://www.lemonde.fr/societe/article/2024/04/30/bac-2024-le-faible-coefficient-affecte-a-l-epreuve-de-philosophie-est-une-anomalie_6230714_3224.html
Bac 2024 : « Le faible coefficient affecté à l’épreuve de philosophie est une anomalie »
Marie Perret, présidente de l’Association des professeurs de philosophie de l’enseignement public, déplore, dans une tribune au « Monde », que la philosophie se voie affecter le plus faible coefficient de toutes les épreuves du baccalauréat, qui « ne rend pas justice au travail » des élèves et des professeurs.
Darren Leathley
Akshar VarmaIn 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. :)
MR.ED Hört Musik#recordoftheday Various - Fuga IV [TOKEN 114]
#stream on https://mredhoertmusik.de/track-of-the-day/token-114-various-fuga-iv/
#mredhoertmusik
#coefficient
#baileyibbs
#phara
#hemka
#nørbak
#NORBAK
#viniciushonorio
#ricardogarduno
#kikepravda
#amorphic
#scifitechno
#hypnotictechno
#techno
#technomusic
#electronicmusic
#bandcamp
