![]()
but go ahead and cling to one outlier poll by quinnipiac. don't say i didn't warn you about your own stupidity and inability to read simple polling data.
I don't think it's an outlier Buck. Maybe. We'll see.
it's one poll that cntradicts what every single other poll says, retard.
![]()
![]()
be stupid somewhere else.
And it was a recent poll.
until other polls confirm its findings, it is an outlier and to be treated as such.
you're welcome for the lesson in math and statistics, courtesy of an ASU dropout.
Could you really pass an actuarial exam Buck?
yes.
tell me again about how nature has intent, which you claimed you learned in college.
Don't you have to hold a degree in order to sit for the exam?
no.
tell me again about how nature has intent, which you claimed you learned in college.
Then why don't you take the actuarial exam and prove you can pass it? I offered to pay for the exam and your time a while ago. The offer still stands. Come to my house, since you know where I live, and I'll give you the money. And feed you some damn fine BBQ and cannabis.
unlike rend pawl, i do not accept money from white supremacists.
i have nothing to prove. go fetch me a sample question and i'll answer it.
unlike rend pawl, i do not accept money from white supremacists.
i have nothing to prove. go fetch me a sample question and i'll answer it.
An insurance company determines that N, the number of claims received in a week, is a random variable with P[N = n] = 1/2n+1, where n > 0 . The company also determines that the number of claims received in a given week is independent of the number of claims received in any other week. Determine the probability that exactly seven claims will be received during a given two week period.
You have 3 minutes to post the answer.
can't you even format the question right so i stand a chance?
you should have typed 1/(2^(n+1)) so i would know what you are talking about.
in which case P[N=7] = summation from 0 to 7 of N1 (week 1) and N2 (week 2).
sum up all the probabilities of week 1 from 0 to 7 then sum up all the probabilities of week 2 using 7-n (because we need exactly 7).
you get the summation from n=0 to 7 of (1/(2^(n+1))) * (1/(2^(8-n)))
the n's cancel out, leaving you with the summation from 0 to 7 of 1/(2^9) = 8/(2^9) = 8/512 = 1/64
excuse me for going out and fetching dinner for my wife and i.
Yeah, you took a long time to look that up on the internet.
And with your search skills, that was an eternity.
i went downtown to get bourbon grill for the wife and i.