Recent content by Skyrior

  1. S

    Short Title:Probability Homework Questions

    Ah, I see. Thanks for the tip. I really appreciate it! (it also makes me wonder why the question is so ambiguous though..) :)
  2. S

    Short Title:Probability Homework Questions

    Thanks so much, for I, I checked the formula for binomcdf and it makes sense now! I also understand IV and II now! Thanks :) (I will ask my teacher about III)
  3. S

    Short Title:Probability Homework Questions

    Homework Statement I. A hundred seeds are planted in ten rows of ten seeds per row. Assuming that each seed independently germinates with probability 1/2, find the probability that the row with the maximum number of germination contains at least 8 seedlings. II. Consider a randomly chosen n...
  4. S

    Finding the sum of a trigonometric series

    erm...I'm quite dumb...I still don't get it. Sum of series of real+imaginary would then be: (1-(a(e^{iθ}))^{n})/(1-(a(e^{iθ})))? Uh... Well, I'm really confused. I know how to get to this part, but I never know how I could remove the imaginary part. The real part of the series is =...
  5. S

    Quadratic Functions Word Problem

    No, you shouldn't. I'm not very sure how to do this in this case, because I only know the Calculus method, sorry
  6. S

    Finding the sum of a trigonometric series

    We didn't learn Euler's Theorem but we learned sum for a finite geometric series. Sorry, but I still don't think I get it. I'm sorry for not stating it, but I did attempt to use geometric series, but I don't know what is r. I think r should be something like a (cos 2θ/cosθ) but (cos 3θ/cos...
  7. S

    Quadratic Functions Word Problem

    Let x be length, y be height: Low Fence = y. 2*8*x = Cost of Horizontal Fence 2*8*y + 4*y = Cost of Vertical Fence + Cost of Low Fence 16x + 20y = 2400 Thus, y = (2400-16x)/20 You might need calculus from here onward. Did you learn calculus?
  8. S

    Finding the sum of a trigonometric series

    1. The problem statement Find the sum of the series: a. 1 + a cos θ + a^{2} cos 2θ + a^{3} cos 3θ + ... + a^{n} cos nθ Apparently, the answer is: \frac{a^{n+1}(a cos nθ - cos(n+1)θ) - a cos θ + 1)}{a^{2} - 2a cos θ + 1} 2. The attempt at a solution = The real part of z^{0} +...
Back
Top