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anonymous

  • one year ago

HELP PLEASE?! VERY DESPERATE!

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  1. anonymous
    • one year ago
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    Id asked this question last night.... but i dont completely understand how they answered it. I'm a bit deperate right now, can soeone please help!

  2. anonymous
    • one year ago
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    ok so for every point i start with, i can make 7 triangles that go over the center

  3. anonymous
    • one year ago
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    if i remove the over countin i get 21 different triangles

  4. anonymous
    • one year ago
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    so the probability should be 21/ 85?

  5. anonymous
    • one year ago
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    sorry, 21/84 = 1/4?

  6. Loser66
    • one year ago
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    |dw:1439602530587:dw|

  7. anonymous
    • one year ago
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    sorry, 21/84 = 1/4?

  8. Loser66
    • one year ago
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    |dw:1439602631083:dw|

  9. Loser66
    • one year ago
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    |dw:1439602655462:dw|

  10. Loser66
    • one year ago
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    |dw:1439602680003:dw|

  11. Loser66
    • one year ago
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    |dw:1439602702369:dw|

  12. Loser66
    • one year ago
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    |dw:1439602734895:dw|

  13. Loser66
    • one year ago
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    |dw:1439602755763:dw|

  14. Loser66
    • one year ago
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    |dw:1439602777272:dw|

  15. Loser66
    • one year ago
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    You see, from AB , you can form 7 triangles but only 1 of them can conclude the center, it it ABF , you have to prove it!! (Fortunately, I did it for you on the previous post)

  16. Loser66
    • one year ago
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    Understand this part?

  17. Loser66
    • one year ago
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    Now, calculate sample space.

  18. Loser66
    • one year ago
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    How many sample space do you have?

  19. Loser66
    • one year ago
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    (OUTCOME)

  20. anonymous
    • one year ago
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    i can create 84 different triangles in total

  21. Loser66
    • one year ago
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    Yes,

  22. Loser66
    • one year ago
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    among them, how many of the triangles include the center?

  23. anonymous
    • one year ago
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    at first I found that for any point, i can make 7 triangles from that point that satisfies the restrictions. Because we can make the the same triangle from three different points, i reduced the overcounting by a factor of three to get 21/84 = 1/4.

  24. anonymous
    • one year ago
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    but now now im getting12 per point, and 12*9/3= 36 different triangles in total. so 36/84

  25. anonymous
    • one year ago
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    because from each point i counted how many cases for each different possible second point., does that make sense? am i thinking right?

  26. anonymous
    • one year ago
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    *sorry i counted how many different triangles i can make that satisfy the restriction for each different second point (which are my cases)

  27. Loser66
    • one year ago
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    ok, you got it.

  28. Loser66
    • one year ago
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    and ganeshie gave you the general form also.

  29. anonymous
    • one year ago
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    yes, but i, not sure how he got it and i need to include an explanation...

  30. Loser66
    • one year ago
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    So, you have to wait for him, hehehe..

  31. anonymous
    • one year ago
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    haha i suppose so

  32. anonymous
    • one year ago
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    thanks so much once again! you're a great teacher!

  33. anonymous
    • one year ago
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    i just need part b. Any help is appreciated :D

  34. Astrophysics
    • one year ago
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    I'll tag @ganeshie8 for you, that way I get to see the post as well hehe ;).

  35. ganeshie8
    • one year ago
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    part \(a\) : number of acute triangle in a regular nonagon is \(30\). Lets start by fixing one vertex, call it \(A\). Then observe that the other two vertices of required triangle must necessarily be on opposite sides of diameter through \(A\) : |dw:1439637980907:dw|

  36. ganeshie8
    • one year ago
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    If you choose the second vertex on right side of semicircle at \(1\) vertex away from \(A\), |dw:1439638852437:dw| then there is exactly \(1\) choice for the third vertex : |dw:1439638864355:dw|

  37. ganeshie8
    • one year ago
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    Similarly, if you choose the second vertex on right side of semicircle at \(2\) vertices away from \(A\), |dw:1439639089591:dw| then there are exactly \(2\) choices for the third vertex : |dw:1439639098201:dw|

  38. ganeshie8
    • one year ago
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    Similarly see if you can count the number of acute triangles with one vertex fixed at \(A\) and the second vertex at \(3\) vertices away from \(A\) : |dw:1439639671473:dw|

  39. ganeshie8
    • one year ago
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    @Emeyluv99

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