Matheus Silva
Reputation
Next privilege 125 Rep.
Vote down
 Sep 26 awarded Notable Question Mar 10 asked Prove by induction that $\frac{x^{n}-1}{x-1}$ is an integer Dec 3 awarded Curious Nov 5 awarded Popular Question Jul 17 accepted Definition of “quotient set” Jun 19 comment Normal distribution of $t$. @Ragnar the total population(10000) the average($\mu$) is 50.03385, and the average from the sampling that i made(i took 10 from 10000) is 70.32416. yes, its correct, i checked using excel. Jun 19 asked Normal distribution of $t$. Jun 19 suggested rejected edit on T student distribution Jun 18 comment (correction)Sampling distribution homework i dont know if i found correctly, based on bernoulli distribution i find the variance = 0,21 and the mean = 0,3, on binomial distribution looking on the histogram and calculatin the mean = 1,5 and variance = 1,05. Which one is correctly? Jun 18 comment (correction)Sampling distribution homework in the question is MSE($\bar{X},\mu$), so, MSE = E($\bar{X} - \mu)^2)$ Jun 18 comment (correction)Sampling distribution homework and to know the squared mean error? Jun 18 accepted (correction)Sampling distribution homework Jun 18 revised (correction)Sampling distribution homework edited title Jun 18 asked (correction)Sampling distribution homework Dec 7 comment identity and inverse elements from a group So, i was thiking better, the inverse is 1/(a,b) = (1/a, 1/b). (a,b)*(1/a, 1/b) = 1 -> a*1/a = 1 | b*1/b = 1 -> (1,1). (a,b)*(j,k) = (1,1) where (j,k) is any inverse, so its the only way to achieve (1,1). Dec 7 comment identity and inverse elements from a group i was thinking better, why i can just consider (0,0) as the inverse? Dec 7 comment identity and inverse elements from a group so, i understand that (C,D) is (1,1) correct? is the only way to achieve (a,b). ok, and considering (0,0) as inverse, multipling by (a,b) (or whatever) it will not exist(will be 0,0) correct? so, dont have an inverse. Dec 7 comment identity and inverse elements from a group @JinyongGo '-', so, how can i show there is not an inverse element? and its not a group Dec 7 comment identity and inverse elements from a group @JinyongGo how it should be? and, the logic is correct? Dec 7 comment identity and inverse elements from a group i wrote the entire question.