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- How do I combine standard deviations of two groups?
I have 2 groups of people I'm working with the data about their age I know the means, the standard deviations and the number of people I don't know the data of each person in the groups Group
- recursive algorithms - Recursion tree T (n) = T (n 3) + T (2n 3) + cn . . .
Recursion tree for T(n) = T(n3) + T(2n3) + cn T (n) = T (n 3) + T (2 n 3) + c n Shortest path will be most left one, because it operates on lowest value, and the most right one will be the longest one, that means tree is not balanced
- Prove that $n! gt; n^{3}$ for every integer $n \\ge 6$ using induction
Prove that n!> n3 n!> n 3 for every integer n ≥ 6 n ≥ 6 using induction [duplicate] Ask Question Asked 6 years, 5 months ago Modified 6 years, 5 months ago
- matrices - Number of Arithmetic Operations in Gaussian-elimination . . .
We know that the total number of multiplications divisions and additions subtractions of the Gaussian-elimination technique is n3 3 + 1 2n2 − 5 6n n 3 3 + 1 2 n 2 5 6 n and 1 3n3 − 1 3n 1 3 n 3 1 3 n respectively
- induction - Prove that $ n^3 + 5n$ is divisible by 6 for all $n\in . . .
So n=3k+1, the remainder of n3 n 3 when divided by 3 is 1 and for 5n 5 n it will be 2 2 and 2 + 1 = 3 2 + 1 = 3 And for n = 3k + 2 n = 3 k + 2, the n3 n 3 leaves remainder 2 when divided by 3 and 5n 5 n leaves remainder 1 1 when divided by 3, hence proved The case is trivial for n=3k
- Proving $n^3$ is even iff $n$ is even - Mathematics Stack Exchange
I am trying to prove the following statement: Prove n3 n 3 is even iff n is even Translated into symbols we have: n3 n 3 is even n n is even Since it's a double implication, I started assuming n is even, then eventually concluded:
- Proof that $n^3+2n$ is divisible by $3$
If n n is divisible by 3 3, then obviously, so is n3 + 2n n 3 + 2 n because you can factor out n n If n n is not divisible by 3 3, it is sufficient to show that n2 + 2 n 2 + 2 is divisible by 3
- divisibility - Proving $n^3 + 3n^2 +2n$ is divisible by $6 . . .
The full question is: Factorise n3 + 3n2 + 2n n 3 + 3 n 2 + 2 n Hence prove that when n n is a positive integer, n3 + 3n2 + 2n n 3 + 3 n 2 + 2 n is always divisible by 6 6 So i factorised and got n(n + 1)(n + 2) n (n + 1) (n + 2) which i think is right? I'm not sure how to actually prove this is divisible by 6 6 though Thanks for help and i apologise if someone has already asked this, i
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