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  1. ... has a BINOMIAL distr. B(n,p) P(N = k) = C(n,k) p^k (1-p)^(n-k) therefore, here : P(N=3) = C(41,3) (1...

    5 Answers · Science & Mathematics · 22/05/2015

  2. ... , assume that (#1) stands when m = k . That is , "(2^k)C(1), (2^k)C(2), ... , (2^k)C(2^k-1) are all even" ---(#2...

    2 Answers · Science & Mathematics · 03/10/2019

  3. .... Probability distribution of x: f(x) = P(x = k) = C(6,k) C(6,3-k) / C(12,3) , k=0,1,2,3 -----(1) Probability...

    2 Answers · Science & Mathematics · 13/07/2012

  4. QE : 1x² + (-k)x + (2k-10) = 0 ∴ a = 1, b = -k, c = 2k - 10 .......... (1) For equal roots, discriminant Δ = b²...

    1 Answers · Science & Mathematics · 09/04/2011

  5. ..., now is just a matter of solving the system. 9 = -3/ k + c ==> c = 9 +3/k Plug that in the second equation. 4 = (1/k) -1 + c...

    4 Answers · Science & Mathematics · 19/12/2010

  6.   This is a linear relationship: C = K − 273 K = C + 273 1. 77°K = 77−273 = −196°C 2. 0°C = 0+273 = 273°K

    1 Answers · Science & Mathematics · 29/04/2016

  7. ...and h. You can see the latter as follows: f(x) = c . a^(d(x-h)) + k = c . a^(dx) . a^(-dh) + k = c' . a^(dx) + k, where c' = c . a^(-dh) So we...

    2 Answers · Science & Mathematics · 06/04/2016

  8. ...prime, we may conclude that p|m. Writing m = cp for some integer c, we get (cp)^k = c^k * p^k = p * n^k ==> n^k = c^k * p^(k-1). Since k > 1, we have...

    1 Answers · Science & Mathematics · 14/10/2009

  9. ...the constant. We know that initially, v(0) = 0, so: 0 = C - g*M/k C = g*M/k v(t) = (g*M/k)*(exp(-k*t/M) - 1) Now, v...

    1 Answers · Science & Mathematics · 28/03/2010

  10. ...the constant. We know that initially, v(0) = 0, so: 0 = C - g*M/k C = g*M/k v(t) = (g*M/k)*(exp(-k*t/M) - 1) Now, v...

    1 Answers · Science & Mathematics · 28/03/2010

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