Answer:
981,518
Step-by-step explanation:
Have a nice day :)
Answer:
200
Step-by-step explanation:
It is given that the same player is going to play a game one after another continuously about 200 times. During the game the player is going to pick up black color balls.
If the player picks at least one black color ball at one time of the game, therefore, the probability of the number of black balls that the player is going to pick is 200 balls.
<h2>
Answer and Explanation to questions 13,14,15</h2>
13)
as given in the question.
14)
Since Y is the midpoint of XZ. So, Y will divide XZ in equal halves into XY and YZ.
15) 
and
. So, 
<h2>
Answer and Explanation to questions 16,17,18</h2>
∠3 is supplementary to ∠1 means: ∠3 + ∠1 = 180°
And, according to figure ∠1 + ∠2 = 180° as ∠1 and ∠2 form a straight line.
∠3 + ∠1 = 180° .............(i)
∠1 + ∠2 = 180° .............(ii)
subtracting equation (i) and (ii) will give ∠3 = ∠2 ..........(iii)
15) ∠3 is supplementary to ∠1 as given in the question
16) ∠2 is supplementary to ∠1 as shown be equation (ii)
18) ∠3 ≅ ∠2 as shown by equation (iii)
<h2>
Answer and Explanation to questions 19</h2>
∠3 and ∠4 form a straight line. Therefore, ∠3 + ∠4 = 180° .......(i)
∠4 and ∠5 form a straight line. Therefore, ∠4 + ∠5 = 180° .......(ii)
subtracting equation (i) and (ii)
∠3 + ∠4 - (∠4 + ∠5) = 180°-(180°)
∠3 + ∠4 - ∠4 - ∠5 = 180°-180°
∠3 - ∠5 = 0
∴ ∠3 = ∠5 (Hence Proved)
Answer:
x = 7.5
Step-by-step explanation:
Since DF bisects ∠CDE then
∠EDF = ∠CDF, hence
8x = 4x + 30 ( subtract 4x from both sides )
4x = 30 ( divide both sides by 4 )
x = 7.5
We have
![\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)](https://tex.z-dn.net/?f=%5Csqrt%5Bk%5D%7B%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%7D%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%5Cright%29%7Dk%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%5Cright%29%2B%5Cln%5Cleft%28%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%5Cright%29%2B%20%5Ccdots%20%2B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%5Cright%29%7Dk%5Cright%29)
and as k goes to ∞, the exponent converges to a definite integral. So the limit is
![\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bk%5Cto%5Cinfty%7D%20%5Csqrt%5Bk%5D%7B%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%7D%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Clim_%7Bk%5Cto%5Cinfty%7D%20%5Cfrac1k%20%5Csum_%7Bi%3D1%7D%5Ek%20%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cfrac%20ik%5Cright%29%5Cright%29%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cint_0%5E1%20%5Cln%5Cleft%28%5CGamma%28x%29%5Cright%29%5C%2C%20dx%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%282%5Cpi%29%7D2%7D%5Cright%29%20%3D%20%5Cboxed%7B%5Csqrt%7B2%5Cpi%7D%7D)