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2 Nation in Peace Mean
What else can I help you with?
If you multiply 2 positive integers will result in what kind of product?
Positive. p*p=p p*n=n n*n=p
What is the formula for solving of induction?
Induction is not a formula, it is a method of proof. Anyway, state the property you wish to prove about each natural number n. This is usually the given P(n). Prove this for the zeroth case, i.e. P(0). Assume the nth case is true, i.e. P(n). Show P(n) => P(n+1). Example: Prove 2 + 4 + ... + 2n = n(n+1) for n >= 0 Proof: P(0) = 0 trivially. Assume: P(n) Show P(n) => P(n+1). 1. 2 + 4 + ... + 2n = n(n+1) 2. 2 + 4 + ... + 2n + 2(n+1) = n(n+1) + 2(n+1) = (n+1)(n+2). QED
A binomial distribution has a mean of 12 and a standard deviation of 2.683 find n and p?
There is not enough information to find n & p. The mean is n*p and the std dev = sqrt (n*p*q). You have to be given n, p or q to have 2 equations 2 unknowns to solve.
How do you use HYPOTHESIS AND TEST STATISTICs?
Means, Proportions and Variance (One population) H_0:μ=μ_0 assuming σ is known z=(x ̅-μ)/(σ⁄√n) N(0,1) NA H_0:μ=μ_0 assuming σ is unknown t=(x ̅-μ)/(s⁄√n) Student t(υ) ν=n-1 H_0:p=p_0 p ̂=x/n or p ̂=(x+2)/(n+4) z=(p ̂-p)/√((p(1-p))/n) N(0,1) NA H_0:σ^2=σ_0^2 u=((n-1) s^2)/σ^2 χ^2 (υ) ν=n-1
How many different ways can you make 12 cents in change?
I'll use these symbols for each coin: P = Penny; D = Dime; N = Nickel 12 P 7 P & 1 N (7 + 5) 2 P & 2 N (2 + 10) 1 D & 2 P (10 + 2)
What does n equal if n plus 2n equals p?
n equals p over 2(like as a division problem. See, you just write n+2n=p then you do this- n+2n=p you take 2 divided by 2 and those cancel ou so if you divide by one so=ide, you have to do it to the other. so you get your answer.
For all integers n if n greater than 2 then there exists prime number p such that n less than p less than n factorial What is the proof?
The proof relies on a result from number theory known as the Bertrand's postulate, which states that for any integer ( n > 1 ), there exists at least one prime ( p ) such that ( n < p < 2n ). Since ( n! ) (n factorial) grows much faster than ( 2n ) for ( n > 2 ), we can conclude that there are primes not only between ( n ) and ( 2n ) but also between ( n ) and ( n! ). Thus, for any integer ( n > 2 ), there exists a prime ( p ) such that ( n < p < n! ).
Formula for Partially ordered set for n elements?
$p(n)\,=\,2^{n^2/4+3n/2+O(\log_2n)}$
How do you get 12.75 with 1 2 3 4?
P(2x3) - 1/4 where P(n) is the n-th prime.
How do you multiply negatives?
P= positive N=negative P x N = N N x P = N P x P = P N x N = P Hope that helps!?!?!
When number N is reduced by p percent the result is then?
N - p% = N - p% of N = N*(1 - p%) = N*(1 - p/100) or N*(100 - p)/100
How do you find the multiplicative inverse of a number mod a number?
Formally, a number n, has an inverse mod p only if p is prime. The inverse of n, mod p, is one of the numbers {0, 1, 2, ... , k-1} such that n*(p-1) = 1 mod p If p is not a prime then: if n is a factor of p then there is no such "inverse"; and if n is not a factor of p then there may be several possible "inverses".
