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Thanks for Helping me for my Report😂❤ (I know this Video was a Year ago)

Thank you

Can you do a video proving/deriving z-scores?

here before david 😋

Here is my proof: Definition QED

Excellent video! Keep it up

Let S be the sample space and let |.| define the cardinality of a set. Let ¢ denote subset for the lack of the actual symbol used in math notation. By definition: For any event A ¢ S: p(A) = |A| / |S| (Or the porportion of the samples in event A divided by the total samples in the sample space) Then: p(A U B) = |A U B| / |S| But note that: |A U B| = |A| + |B| - |A ^ B| where ^ here denotes intersection. This is true since if the sets were not disjoint, then by adding their cardinalities we would have counted some elements twice so we need to remove away the ones we over counted and those are exactly the elements that occur in both. Thus: p(A U B) = (|A| + |B| - |A ^ B|) / |S| = |A| / |S| + |B| / |S| - |A ^ B| / |S| = p(A) + p(B) - p(A ^ B)

In my class, we have to use generating function to prove this

I don't know why TH-cam recommended me this video, but nice video man. It remebers me when I used to have statistics classes in college.

How it feels to chew 5 gum

Thaks from indai

what course is this?

David missed a banger… 😢

P is a measure, measures are stable by complementarity, and by definition of a tribe and of a probability measure, P(omega)=1, therefore P(Null)=P(omega^c)=P(omega\omega)=P(omega)-P(omega)=0

So this means that something always happens?

So the probability that nothing happens is null?

Compound angle formula can prove the formula of multiple and submultiple angle.

Back to back bangers 🐻🔥

Another banger 🔥🔥🔥

Sir very very thank u for imp message or formula 🎉 thank u

Thank you for sharing ☺️

Nice

Great, Thanks

Can we prove it by starting with the LHS, with only the 12 properties of numbers?

Thanks man .u save mee

Ive heard u somewhere ,u feel a little like technoblde

Why is ax+bcy=1 imply gcd(a,bc)=1

kkk took my baby away 🤧

mr. snell my GOAT released another banger video

a(a²-1)=(a-1)a(a+1)... a product of 3 consecutive numbers; one of them is guaranteed to be a multiple of 3 now consider that a is odd... which means both a-1 and a+1 are even; Furthermore, since a=4n+1 or a=4n-1, either a-1 or a+1 are a multiple of 4; So one is a multiple of 4, and the other is a multiple of 2... hence the product will be at least a multiple of 8; Since the a(a²-1) is a multiple of 3 and of 8.... it must be a multiple of the least common multiple between the 3 and 8, which is 24.

Well done. It seems that this could be generally true for any multiple 6*n-1.. so: 6 | (a^3 + a*(6*n-1)). If thats the case, is it further generalized to other coprime factorizations with 2 factors - not just 6? like 10, 14, 15, 21, 22, etc?

Ooo I love this one!

are you sure this is correct? your father is asking 🤓

FIREEEE KEEP POSTING WE MISS YOU

Full support

Watching this again. Thanks for making it easy to understand

just represent a and b as products of primes by the fundamental theorem of arithmetic. Bezout is rarely used in competitive maths since it can often be avoided, but still an awesome solution.

i have acquired you two new subs ☝️

I am from India 🇮🇳

Thanks sir

FIRST

You have just light up my day

Nice proof.

first 😮

Sir please continue.love from India 🎉

Thank you so very much for explaining these proofs. This most certainly helped.

Excellent. Thanks so much or taking the time to explain. I appreciate this.

YAYAYAY NEW SNELL BROS MATH VIDEO

Thanks for this explanation. Can you do more proofs in elementary number theory? Probably using the Burton textbook. Thanks

Thanks for this video. It helped.