anonymous
  • anonymous
HELP PLEASE?! VERY DESPERATE!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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!
anonymous
  • anonymous
ok so for every point i start with, i can make 7 triangles that go over the center
anonymous
  • anonymous
if i remove the over countin i get 21 different triangles

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anonymous
  • anonymous
so the probability should be 21/ 85?
anonymous
  • anonymous
sorry, 21/84 = 1/4?
Loser66
  • Loser66
|dw:1439602530587:dw|
anonymous
  • anonymous
sorry, 21/84 = 1/4?
Loser66
  • Loser66
|dw:1439602631083:dw|
Loser66
  • Loser66
|dw:1439602655462:dw|
Loser66
  • Loser66
|dw:1439602680003:dw|
Loser66
  • Loser66
|dw:1439602702369:dw|
Loser66
  • Loser66
|dw:1439602734895:dw|
Loser66
  • Loser66
|dw:1439602755763:dw|
Loser66
  • Loser66
|dw:1439602777272:dw|
Loser66
  • Loser66
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)
Loser66
  • Loser66
Understand this part?
Loser66
  • Loser66
Now, calculate sample space.
Loser66
  • Loser66
How many sample space do you have?
Loser66
  • Loser66
(OUTCOME)
anonymous
  • anonymous
i can create 84 different triangles in total
Loser66
  • Loser66
Yes,
Loser66
  • Loser66
among them, how many of the triangles include the center?
anonymous
  • anonymous
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.
anonymous
  • anonymous
but now now im getting12 per point, and 12*9/3= 36 different triangles in total. so 36/84
anonymous
  • anonymous
because from each point i counted how many cases for each different possible second point., does that make sense? am i thinking right?
anonymous
  • anonymous
*sorry i counted how many different triangles i can make that satisfy the restriction for each different second point (which are my cases)
Loser66
  • Loser66
ok, you got it.
Loser66
  • Loser66
and ganeshie gave you the general form also.
anonymous
  • anonymous
yes, but i, not sure how he got it and i need to include an explanation...
Loser66
  • Loser66
So, you have to wait for him, hehehe..
anonymous
  • anonymous
haha i suppose so
anonymous
  • anonymous
thanks so much once again! you're a great teacher!
anonymous
  • anonymous
i just need part b. Any help is appreciated :D
Astrophysics
  • Astrophysics
I'll tag @ganeshie8 for you, that way I get to see the post as well hehe ;).
ganeshie8
  • ganeshie8
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|
ganeshie8
  • ganeshie8
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|
ganeshie8
  • ganeshie8
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|
ganeshie8
  • ganeshie8
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|
ganeshie8
  • ganeshie8
@Emeyluv99

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