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- ...e the process of this demonstration due to the limited environment to draw integral. That integral calculation might be tough one, but it would not be a big deal.1 KB (248 words) - 21:14, 4 October 2008
- ...Then to find the PDF of the whole chord, i just used the formula with the integral and used the parameters of D for the limits and fx(x) as 2 times the L.513 B (104 words) - 13:45, 6 October 2008
- ...2*sqrt(r^2-D^2). Next i said Fsub(X)(x)= L= 2*sqrt(r^2-D^2) and take its integral from 0 to 2r. This is just a thought dont know if its correct.382 B (79 words) - 18:08, 6 October 2008
- because we are doing an integral of x, and the probability that x < y or x > 1 is 0, the limits of integrati1 KB (228 words) - 13:23, 22 November 2011
- In other words, remember that the integral over all Y for every PDF must equal 1. So, since you know that Y must now b701 B (129 words) - 18:03, 15 October 2008
- ** Okie, E[y] = the integral from -inf to +inf of (v*f(v) dv)306 B (55 words) - 17:19, 16 October 2008
- ...be a possibility a corresponding x-coordinate is NOT in the triangle, the integral becomes:<br />1,016 B (166 words) - 13:27, 22 November 2011
- take the integral from an integer k-1 to k of the function lambda*X*e^(-lambda*X) and that is213 B (45 words) - 06:37, 17 October 2008
- P[H1] = integral from 0 to 1 of P(H1|Q=q)fQ(q)dq = integral from 0 to 1 of q(2q)dq194 B (46 words) - 09:55, 20 October 2008
- P[H1] = Integral from 0 to 1 q(2q)dq P(H2 n H1) = Integral from 0 to 1 q^2(2q)dq204 B (46 words) - 12:59, 20 October 2008
- Fu(U) = P[U<= u) = integral from -inf to +inf of 1 du = u120 B (29 words) - 16:51, 20 October 2008
- *i.e. the first integral will look something like this: <math> f_z(z)= \int \limits_{0}^{\infty} \la2 KB (344 words) - 17:00, 21 October 2008
- Another integral to convolute is <math> f_z(z)= \int \limits_{z}^{\infty} \lambda e^{-\lambd196 B (37 words) - 18:58, 21 October 2008
- *E[1/x] = integral(<math>\lambda e^{-\lambda x}</math> * (1/x)) dx * this integral is undefined182 B (28 words) - 14:48, 10 November 2008
- I would suggest splitting the double integral up. (Think of a double integral as a nested for loop -- integrating "slowly" over the outside loop and "qui1 KB (167 words) - 18:33, 9 December 2008
- E[x-q(x))^2] = Integral from -inf to inf (x-q(x))^2*fx(x)dx =integral from 0 to 1 (x-q(x))^2dx253 B (48 words) - 08:44, 10 December 2008
- Riemann Sum for the integral719 B (133 words) - 10:49, 14 October 2008
- Evaluate the Integral:1 KB (259 words) - 08:19, 1 October 2008
- Evaluate the integral: Good work. That last integral is easier to look at if you write <math>e^{-x}</math> in place of <math>\fr1 KB (260 words) - 07:50, 3 October 2008
- ...a <math> \frac{t}{p} </math> so I would have a ''dt''. That led me to the integral below. Does it make sense and does anyone know how to integrate the proble I don't know how to use this integral, but I did some manipulation and got this:1 KB (270 words) - 09:43, 7 October 2008