you are given the standard form z = 3 - 3i
The Attempt at a Solution
so to convert this to polar form, i know that ##r = 3√2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in...
Sigh it's just annoying having to deal with something i did several years ago in a calculus course. I guess I should've remembered the basic factoring property.
Anyway, $$sin^2x + cos^2x - 2sinxcosx = 1/9$$
and I think $$sin^2x + cos^2x = 1$$ so I believe it'll end up being after...
$$sinx - cosx = 1/3$$
solve for $$sin(2x)$$
$$sin^2x + cos^2x = 1$$
$$sin2x = 2cosxsinx$$
The Attempt at a Solution
I think you can square both sides and get:
$$sin^2x - cos^2x = 1/9$$
But how can I use this information to solve for sin2x? Is there a...
My calc class is having me review precalc(which I'm really rusty on...)
21. Find sin θ, sec θ, and cot θ if tan θ = 27
22. Find sin θ, cos θ, and sec θ if cot θ = 4.
23. Find cos 2θ if sin θ = 15
24. Find sin 2θ and cos 2θ if tan θ = √2
25. Find cos θ and tan θ if sin θ =...
Yes you would have to restart if you fail on your first attempt on trial 2. Your 9 lives continue from trial 1.
And let's say I'm on the third trial - however, the 'gamemaster' decides to let me have 2 shots at trying to win trial 3 rather than having me restart. This would increase the joint...
Yes I believe that you have the right idea here. So if you fail at any instance, you have to restart. Once you have used up your 15 attempts, you lose.
So in essence, if allowed a second attempt at the same trial(just for the sake of understanding) then essentially the joint probability is...
I have another thread where there's a game similar to this one(however simpler to some degree) and I learned quite a bit from it. However, my neighbor had designed a game when he was in university for a project, and I would like to understand a bit more about it. .
So that would be around 40 correct?
I had gotten a .108 as the answer to the expression you had given, and that yields a 89% of failing one attempt.
So of which around 40 tries would have been sufficient?
I have a much more complicated game that I'd like to ask how to accurately formulate...
Right, I'd like to find the minimum amount of players required to essentially have a 99% of one of them winning the game. But just for the sake of rules here, if no one wins then those two trials then that's that.
I re-read my original post and I can see why it's a little confusing here. I'll...
I apologize for the lack of responses here on my part, I waas quite busy with school work. If you don't mind bringing this topic back up, I would like to clarify some things.
You're right about me not explaining well here. Originally I had the game set up where you could try endlessly until you...
This is extremely helpful. Knowing more information about the trials is actually interesting, because I was thinking a bit further on how the markov model may be needed more.
I was thinking about adding another element to this game that might make things a little more complicated. Let's say...
I'm doing a bit of research on Markov's model as of the moment, I'm trying to understand it through some youtube videos and I can see where it branches off. However the method mentioned above seems to be very straight to the point.
So... it would take about 63 attempts to actually succeed all 3...
Right now what I want to figure out is the most optimal way of winning all three tries with the least amount of attempts. However, there are no outside factors influencing success and failure, they are merely chances predetermined. That is why I am trying to figure out how many attempts it would...
I'm all ears for learning this, however, I have no background in any form for probability. I'd be surprised if any of this was not alienated to me. Just a question though, using this more complex model, i imagine there would be a more accurate answer obtained in comparison to the methods used above?
This isn't a homework question really... it's just something I'm not sure how to tackle.
Say you have a 44% chance of being successful on one try, and then another 44% to succeed on your second try, and then a 36% to succeed on your final try.
If you fail once at any...