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Use the binomial probability formula: P(X =

**k**) =**C**(n,k) * p^k * q^(n-k) n : number of trials (5) k : number...2 Answers · Science & Mathematics · 12/01/2017

Use the binomial probability formula. P(X =

**k**) =**C**(n,k) * p^k * q^(n-k) n : number of trials (10) k : number...3 Answers · Science & Mathematics · 31/12/2016

... to fix the unity element? i.e. Is ϕ(1_F) = 1_

**K**? (**c**) Is {0} a subring of Q? If you said (a) yes; (b) yes...1 Answers · Science & Mathematics · 27/07/2016

...have ln(e^5) = k*2 + C. So we get ln(e^5) is 5, and 5 = 2*

**k**+**C**. With y(5) = e^2, we get 2 = 5***k**+**C**. Solving the 2 ...2 Answers · Science & Mathematics · 24/07/2016

Reinsert your brain properly, kid. You'll be more coherent, then.

4 Answers · Entertainment & Music · 22/06/2016

You want us to do that without first defining a,

**k**,**c**, or d? Followup: Your definitions answer your question...1 Answers · Science & Mathematics · 22/09/2018

This is a linear relationship: C = K − 273

**K**=**C**+ 273 1. 77°K = 77−273 = −196°C 2. 0°C = 0+273 = 273°K1 Answers · Science & Mathematics · 29/04/2016

...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

...be considered a double application of Pascal's rule: C(n,

**k**) =**C**(n-1,k-1) + C(n-1,**k**) =**C**(n-2,k-2) + C(n-2,k-1) + C...2 Answers · Science & Mathematics · 28/03/2016

=> f(x) = x^2 - kx + 5 a = 1 b = -

**k****c**= 5 b^2 - 4ac = 0 => only one zero. y^2 - 4*1*5 = 0 y^2 = 20 y = ±2√51 Answers · Education & Reference · 21/03/2016

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