So... a student sent me a bunch of word problems. The thing is that in those word problems the protagonists are jerks to one another.

 

Makes me wonder about the mental state of the math instructor who produced those word problems.

 

#math #WordProblems #jerks

Must've killed off a few hundred thousand brain cells last night, because I was trying to come up with a particular word, and the only thing that deigned to exit my mouth in **multiple** tries was "skeleton of a horse's head."

Horse skull. I was trying to say horse skull.

#language
#vocabulary
#WordProblems

Apparently AI doesn’t like word problems either:

“The research paper explains that the best and brightest LLM models saw ‘catastrophic performance drops’ when trying to answer simple math problems that were written out like this.

It happened primarily when those problems included irrelevant data, which even schoolchildren quickly learn to disregard.”

#ArtificalIntelligence #AppleIntelligence #LargeLanguageModels #Technology #Math #WordProblems #Data #Research

https://9to5mac.com/2024/11/01/apple-researchers-ran-an-ai-test-that-exposed-a-fundamental-intelligence-flaw/

Apple researchers ran an AI test that exposed a fundamental ‘intelligence’ flaw

Apple’s AI research recently produced a paper outlining a fundamental flaw in AI’s ‘intelligence,’ even as the company goes all-in on AI.

9to5Mac

It's kind of hard to calculate this directly but of course there is a trick! We're going to calculate the chance that neither cop is a domestic abuser instead. Then all other sets of two cops will contain at least one

There's a 60% chance that cop one isn't and a 60% chance that cop two isn't. This makes the chance that both aren't 0.6 x 0.6 = 0.36. so 36% of pairs of cops fail to contain a domestic abuser, which means 64% contain at least one. If you see two cops and say that there's a 64% chance that one is an abuser, well, that is correct!

This keeps working! The chance that none in a set of three is an abuser is 0.6 x 0.6 x 0.6 = (0.6)^3 = 0.216, so there's a 78.4% = 100-21.6 chance that a set of three cops contains at least one abuser.

If there are n cops the formula is:

(1-(0.6)^n)*100

n. % chance of at least one
1. 40
2. 64
3. 78.4
4. 87
5. 92.2
6. 95.3

Word problems don't have to be counterrevolutionary...

#ACAB #PoliceAbolition #Abolition #WordProblems #DiscreteProbability #Mathstodon

If your nephew sold 27 old skateboards for 300 dollars, what color shoes is he wearing? #word_problems