u have a pol on for those angles,
this is why tangens isnt defined here,
but still u can consider those points as limits of sequences which goes ins unlimited
i dont always try to explain it a very correct way, even if my explanation have the right part.
This is why i had ;) at the end.
My last answer was corrrect.
Pol is a place, where to the point belongs an unlimited big number.
For example 1/|x| has a pol on 0, coz there the function has the number unlimited.
(dunno if "pol" the right word in english, at least its in german so :) )
ok here a task: imagine u ar in the forest and u will be attacked by 4 wolfs.
Assume u have 5 cartridges (means u can do only 5 shots), and ur shooting accuracy is 80 %.
1. Whats the probability, to kill the last wolf with the last cartridge?
2. Whats the probability to survive the attack of wolfes?
1. Apparently we assume that u have killed the other wolves and have 1 wolf and 1 round left, then it's 80% (Or do u mean the probability that we get into a situation where u got 1 wolf and 1 round left?)
2. 1 - (0,2 + 0,8*0,2 + 0,8*0,8*0,2 + 0,8*0,8*0,8*0,2) = 0.41 (or just 0,8^4 = 0.41)
dunno if i understand u correctly, u should assume, that u can shoot 5 times, where ur accuracy to hit in every shot is 80 %, and u have to kill the 4th wolf with last (means 5th) cartridge
lol :) u can kill all wolfes with 4 shoots of course, but if u fail by anyone shot, u have only 5th last cartridge for killing the 4th wolf
it was meant u can shoot maximal 5 times. If u hit is another question
1/5 you miss the first shot, 1/25 you miss the second shot as well, 1/125 you miss the third shot as well, 1/625, you miss the 4th shot as well, 1/5^5, you miss the last shot as well
imagine u try to arrive a callcenter. The probability to arrive a callcenter of every of ur trials is 0.6.
How much trials do u need in the average to arrive a callcenter?
yes, r should be 1 coz of ur situation, and this is exactly the geometrical distribution.
Ok u have 1-p/p. This gives u the expectation/average of the number of the trials before u get the call. And that means, the expectation is at 1/p=1,6667th call
==> the average arrive number is 1,6667
So you actually visited the 4th grade.
Nice to know
I'm off
tan 100+k200 = ERROR (in gra)
tan pi/2+kpi = ERROR (in rad)
K is an element of Z
this is why tangens isnt defined here,
but still u can consider those points as limits of sequences which goes ins unlimited
Also, "coz its very very big then ;)" simply is a wrong explanation.
This is why i had ;) at the end.
My last answer was corrrect.
Pol is a place, where to the point belongs an unlimited big number.
For example 1/|x| has a pol on 0, coz there the function has the number unlimited.
(dunno if "pol" the right word in english, at least its in german so :) )
e: without wikipedia :<
Assume u have 5 cartridges (means u can do only 5 shots), and ur shooting accuracy is 80 %.
1. Whats the probability, to kill the last wolf with the last cartridge?
2. Whats the probability to survive the attack of wolfes?
2. 1 - (0,2 + 0,8*0,2 + 0,8*0,8*0,2 + 0,8*0,8*0,8*0,2) = 0.41 (or just 0,8^4 = 0.41)
but u are right I should add there like 0,8^4 + 4* 0,8^4*0,2 = 0,74
because I can miss one bullet
it was meant u can shoot maximal 5 times. If u hit is another question
I could do still one more now, then I'll study for my exam :p
Ill post one now, dunno if u will able to solve this, sec
ok, I'll check the exercise now
the task is very easy if u know what is a negative binomial distribution
So , it's 4^5/5^5 ?
aka 0.33
4 shots out of 5 that you hit him each time right ?
80%
so you have 5 shots , hence power 5
np
solve this:
INTEGRAL[e^t * sin(t) - 2t * cos(t)]*dt over 0 to 2 Pi
I won't bore u with the details.
but I'll look into it if I got time... the first term can be Derived from 1/2*e^t(sint-cost) but haven't figured the 2nd one yet :p
it even gives the right answer when u insert the values:
F(t) = 0,5*e^t(sint-cost) - 2*(t sint + cost)
F(2Pi) - F(0) = -267,24583
what do u study btw?
ye i know i also have harder ones, but that one made me sweat last exam cuz i failed and had to do it 3 times :DD
took a fucking hour of testtime :DD
yeah I did it!
How much trials do u need in the average to arrive a callcenter?
If I understand it right I should use this formula?
What do u have to use for r?
Ok u have 1-p/p. This gives u the expectation/average of the number of the trials before u get the call. And that means, the expectation is at 1/p=1,6667th call
==> the average arrive number is 1,6667
ok i go sleep bb
I go study for my exam now :P
0^0=
inf^0=
inf/inf=
0/0=
0*inf=
1^inf=
hf
0
1
uncertainty
uncertainty
0
1
And there ar no limites for using del'hospital
1^x = 1 for x = 1
Assuming 1^n = 1. Need to prove it for 1^(n+1)
1^(n+1) = (1^n)*1 = 1^n which by our assumption is 1.
This means that 1^x = 1 for all x.
QED
But hey, at least it holds for all natural numbers!