Answer:
yes
Step-by-step explanation:
:D
I think it’s the triangle
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)
<span>0.053
The standard deviation of a binomial variable is
sqrt(n*P*(1-P))
where n is the sample size and P is the probability of one of the outcomes.
The sample size is 55 and we'll assume that both male and female are equally likely outcomes.
standard deviation = sqrt(55*.5*.5) = 3.7081%
44% is 1.6181 standard deviations away from the mean of 50%.
Looking this up on a standard deviation chart gives .0526 probability which we'll round to 0.053</span>
Hello!
We have
x+4/2x-1<0
x+4<2x-1
x>5 Hope this helps you out. Any questions please just ask. Thanks so much!
