HELP PLEASE?! VERY DESPERATE!

- anonymous

HELP PLEASE?! VERY DESPERATE!

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- 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

ok so for every point i start with, i can make 7 triangles that go over the center

- anonymous

if i remove the over countin i get 21 different triangles

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## More answers

- anonymous

so the probability should be 21/ 85?

- anonymous

sorry, 21/84 = 1/4?

- Loser66

|dw:1439602530587:dw|

- anonymous

sorry, 21/84 = 1/4?

- Loser66

|dw:1439602631083:dw|

- Loser66

|dw:1439602655462:dw|

- Loser66

|dw:1439602680003:dw|

- Loser66

|dw:1439602702369:dw|

- Loser66

|dw:1439602734895:dw|

- Loser66

|dw:1439602755763:dw|

- Loser66

|dw:1439602777272:dw|

- 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

Understand this part?

- Loser66

Now, calculate sample space.

- Loser66

How many sample space do you have?

- Loser66

(OUTCOME)

- anonymous

i can create 84 different triangles in total

- Loser66

Yes,

- Loser66

among them, how many of the triangles include the center?

- 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

but now now im getting12 per point, and 12*9/3= 36 different triangles in total. so 36/84

- 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

*sorry i counted how many different triangles i can make that satisfy the restriction for each different second point (which are my cases)

- Loser66

ok, you got it.

- Loser66

and ganeshie gave you the general form also.

- anonymous

yes, but i, not sure how he got it and i need to include an explanation...

- Loser66

So, you have to wait for him, hehehe..

- anonymous

haha i suppose so

- anonymous

thanks so much once again! you're a great teacher!

- anonymous

i just need part b. Any help is appreciated :D

- Astrophysics

I'll tag @ganeshie8 for you, that way I get to see the post as well hehe ;).

- 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

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

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

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

@Emeyluv99

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