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Everything posted by CanadaAotS
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Wait so [math]-z*\theta_{error} \leq \mu \leq z*\theta_{error}[/math] [math]P(-z < Z < z) = .96[/math] [math]P(Z < z) - P(Z < -z) = .96[/math] [math]P(Z < z) = .98, P(Z < -z) = .02[/math] [math]z = 2.05[/math] [math]-2.05*8.7517 \leq \mu \leq 2.05*8.7517[/math] OH! I GET IT NOW! [math]-24 \leq \mu_x \leq +24[/math] So, if I divde away 2.05 from 24 (the z value to get .96 confidence) [math]-11.7073 \leq \mu_x \leq 11.7073 [/math] I plunk that into the equation you gave me... [math]-11.7073 = \frac{71}{\sqrt{n}}[/math] Solve [math]n = (\frac{71}{-11.7073})^2[/math] Giving me [math]n = 36.779[/math] THEREFORE I need an SRS of size 36.779 (app) to predict [math]\mu[/math] within .96 confidence. YAY! I think its funny when I learn the material the day before the assignment is due (I'm the worst procrastinator in the world btw)
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Lets see if I have this right "... time per week studying for a first year student follows a normal distribution with a standard deviation of 70.1 minutes." What sample size is required to estimate the true mean weekly study time to within 24 minutes with 96% confidence? This is an actual question on the assignment. So... [math]24 = \frac{71}{\sqrt{n}}[/math] Would be the equation, right? Solved its: [math]\sqrt{n} = \frac{71}{24}[/math] [math]n = (\frac{71}{24})^2[/math] which is equivalent to 8.751736111... Is this what I'm looking for? lmao. I'm still confused EDIT: so does this mean that an SRS of size 8.751736111... would accurately predict the mean within 24 minutes at a .96 confidence?
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Mokele... you have no idea how much you just helped me :p I can now do an entire assignment (due tomorrow btw)
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hijacked ted hijack read jack read thed damn. apparently not! I clearly see navy and black diagonal lines... however if I stare at it for awhile it becomes black. Some kinda optical illusion... O.o
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Don't think of a photon as a particle then simple as that. Think of it as a 'packet' of energy, since thats what it is pretty much. When light isn't interacting with anything, the energy acts like a wave. When it does interact with something it acts like a 'packet'. Hope that helps in your visualization.
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In the forums defence, we try to tell them their wrong first, but they then proceed to proving themselves idiots even more. Its a viscous cycle
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*gives Atheist an award for "Best Cause of Gravity Theory* lol All we have to do is knock the little green gremlins out, and woohoo! anti-gravity is within our grasp!
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How did you find this site?
CanadaAotS replied to blike's topic in Suggestions, Comments and Support
I think it might have been a link from Discover.com... or something. I dont remember exactly lol -
Yah. It's like saying, "well light heats us up. Its has different frequencies. yada yada yada. Why would it need a photon?" Its not so much whether we need it or not. It's about trying to explain the universe, and how it actually works. We could probably think up some really funny explanations about why things are the way they are, but they wouldn't necessarily be true. I hope at least some of what I said has any relevance lol.
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I really need help here lol. I have an assignment due this friday, the whole thing is on confidence intervals, and I have no clue what I'm doing -_-' OK! So, what I do know is that a confidence interval is used to test how sure you are that a predicted SRS of size n has a certain mean. How would I do an example question like We know some statistic has standard deviation of 5.00 (nice and easy lol) If we wanted to know how big of an SRS is needed to predict the mean within a standard error of 2.00 with .95 confidence, how would I do that? I know if I had even a simple question like this explained to me, I'd be able to tackle any of the harder questions. This is how far I can get: [math]M - z * \frac{\theta}{\sqrt{n}} \leq \mu \leq M + z * \frac{\theta}{\sqrt{n}}[/math] Where [math]M[/math] is the sample mean, [math]\theta[/math] is the standard deviation, [math]\mu[/math] is the mean and [math]n[/math] is the SRS size. For the above example, I get [math]M - z * \frac{5}{\sqrt{n}} \leq \mu \leq M + z * \frac{5}{\sqrt{n}}[/math] I believe P(-z<Z<z) = .95, which would make z = 1.96 [math]M - 1.96 * \frac{5}{\sqrt{n}} \leq \mu \leq M + 1.96 * \frac{5}{\sqrt{n}}[/math] And thats as far as it goes. I'm stuck with that damn [math]M[/math] lol. I know I'm supposed to be solving for n... Thanks in advance! ~CanadaAotS
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By the way, the zeros x = 0 and x = -1 are vertical asymptotes. EDIT: oops didnt see page 3 of this thread. This imformation is a tad late
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Could use help with a math problem please.
CanadaAotS replied to slickinfinit's topic in Mathematics
Asking him why theres no dinosaurs in the bible: not subtle rofl -
[math]f'(x) = \frac{-20x^4 + 60x^2 + 40x}{(x^3 + x^2)^2}[/math] So, you got f'(x) and you want f'(x) = 0? This doesn't seem to have any problems lol [math]0 = -20x^4 + 60x^2 + 40x[/math] [math]0 = -20x(x^3 - 3x - 2)[/math] For starters there would've been a local extrema at [math]x = 0[/math] but [math]x \not= 0[/math]... you say x = 2 is one, but since this is an equation of degree 4, there will be 4 zero's. [math](-20x)(x^3 - 3x - 2) = 0[/math] [math](-20x)(x - 2)(\frac{x^3 - 3x - 2}{x - 2}) = 0[/math] [math](-20x)(x - 2)(x^2 + 2x - 1) = 0[/math] [math](-20x)(x - 2)(x + \frac{2 \pm \sqrt{8}}{2}) = 0[/math] [math](-20x)(x - 2)(x + 1 \pm \sqrt{2}) = 0[/math] The roots of f'(x) are [math]0, 2, 1 + \sqrt{2}, 1 - \sqrt{2}[/math], and local extrema's are located at all of the roots except 0. By the way, on step 2 -> 3 I did binomial long division to divide x - 2 into x^3 -3x - 2. I also used the quadratic formula in the last 2 steps. Hope that helps EDIT: btw tree, awesome pi site in your sig! lol...
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Hmm... You sparked my curiosity about july 2002! [math]t=-2.5[/math] [math]P(t) = \frac{20(4t + 3)}{2t + 5}[/math] [math]P(-2.5) = \frac{20(-7)}{0}[/math] Ooooo... I see, its a asymptote of some kind ... I assume it goes to negative infinity...
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ok... its been awhile since I've done this stuff but... 1 lbs = .454 kg mass = 40.86 kg Fg = 40.86*9.8 Fg = 400.428 Ff = mu*Fg Ff = .9*400.428 Ff = 360.3852 Since f = ma 360.3852 = 40.86a a = 8.82 m/s Do I win a cookie?!? lol...
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This is completely incorrect
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For anyone who doesnt know what pascal's pyramid is: 1 1,1 1,2,1 1,3,3,1 1,4,6,4,1 1,5,10,10,5,1 etc. So I wanted to make an excel sheet, where the two values diagonally above add up for the value underneath. Is there any easy way of doing this? Something like, all values selected are calculated by value 1 above and to the left plus value 1 above and to the right... or something along those lines. This would be so much easier then doing it for each box seperately.
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I completely agree with the_simpsons. An organism that gets nutrition from drinking blood I would consider a vampire... Hey, whats the nutritional value in blood for a human anyway? lol And I know a girl whos allergic to everything heh, that would include sun. So, if she was allergic to garlic as well, and just so happened to have sharper canines then normal AND drank blood... Would that make her a vampire? (by Azure's definition)
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Whats allopatry?
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omg0dz its gonna be sci-fi war! *runs into his stargate and closes the iris behind him* lol... but yah, there is no such thing as infinite density, curvature, etc. thats what a mathematical singularity spits out. I like to think of the centre of a blackhole like a 1/0 in a math equation
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Spirits => Utter Crap Demonic Thing-A-Mah-Jigs => Also Utter Crap Now, if you want to be at all scientific about this... I wouldn't suggest trying this if you actually believed it. It would be very easy to scare yourself (along with the spinning and chanting) into a trance state where it would be very possible to hallucinate "bloody mary" in the mirror. However in Real Life™ 'evil spirits' do not pop out of mirrors to kill / maim the un-wary EDIT:: Unless its Sayanora in your mirror
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"Singularities" arent blackholes... singularities are points in mathematical equations where the said equations stop making sense. Thats why blackholes and the big bang as 2 examples are seen as singularities... however a singularity is NOT a physical object of any kind, just a mathematical problem... People seem to confuse singularities with actually things alot
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"supernatural ablities really just higher dimensonal perceptions?"
CanadaAotS replied to darkangel199's topic in Speculations
Wow... that must be one Helluva-boring job you got there to cook this up while staring at a computer screen lmao -
I love the hammer analogy: "Manufactured artifacts are isolated. You make one hammer. Then you make another. But in biological organisms you have the equivalent of one hammer making another hammer. The hammers are connected by ancestor-descent." Now, how often do you get to say "The hammers are connected by ancestor-descent"