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#26 Re: Help Me ! » equilateral triangle in a square » 2012-03-17 00:48:37

I can't understand how "=(" has been replaced by a smiley

#27 Re: Help Me ! » equilateral triangle in a square » 2012-03-17 00:44:56

I tried to calculate the side of an equilateral triangle inscribed in a square in the general case. I refer to your figure, assume BD=a,  BG=d, BH=x and DI=y.  Pythagore theorem applied to GH, GI and HI as hypothenuses gives the two equations :
x^2+d^2 = (a-d)^2+y^2       (1)
x^2+d^2 = a^2+(y-x)^2       (2)
from which I should calculate x and y when GHI is equilateral. I thought that was imposssible, which explains why I found no refernce to this problem on Internet. Later I took a different approach to the problem and found very easily the solution
x= (2*a-d)/sqr(3) and y= (a+d)/sqr(3)
I wonder how we can get these values directly from equations (1) and (2)

#28 Re: Help Me ! » Question on BB Code on this forum » 2012-03-16 09:48:09

Thank you very much ! I have another problem : when I try to add an image with the "img" code the forum responds that I have to be an "established member". How can I be an established member ?

#29 Help Me ! » Question on BB Code on this forum » 2012-03-16 09:25:39

hammana
Replies: 3

Hello,

I see a lot of people hiding their answers in their posts, so that you have to click to display the answer. I have tried to do it using the "spoiler" tag, but it does not work. Which tag should I use to hide my answers too ?

See this test :
[spoiler]it does nothing[/spoiler]

#30 Re: Help Me ! » trigonometry and tangent problem, please help. » 2012-03-14 09:30:18

if the cos(x) is negative then x>pi/2
I think that 2y+4=pi-0.841

#31 Exercises » Probility with two dice » 2012-03-13 10:24:22

hammana
Replies: 1

a - A pair of dice is thrown 11 times. What is the probabiliy that the eleven possible values  of their sum are obtained once each (this can be solved by simple probability theory).

b - if the pair of dice is thrown 12 times, what is the probability that the eleven possible values  of their sum are obtained at least once each

#32 Re: Exercises » Sums of series with odd positive denominators » 2012-03-11 09:37:22

For any positive a and b we have
1/(a-b)+1/(a+b)>2/a
therefore
1/(25-4)+1/(25+4)>2/25 
1/(25-2)+1/(25+2)>2/25

#33 Re: Help Me ! » Little oh help!!! » 2012-03-08 01:25:11

aproximate f(x) by ax+b, calculate the two most significant terms of (ax+b)^4
coefficient of x^4  =  a^4 should equal 1
coeficient of x^3=2* b*a^3 should equal 8
therefore a=1, b=2
It means that as x increses the fuction f(x) aproaches the straitht line y=x+2
To know whether it is above or below, you should use y=ax+b+c/x
It happens in this case that c=0, we must go to y=ax+b+d/x*x
I find d=-2 (assuming  my calculations are right) which means that f(x) is below y=x+2 for x positive
and above for x negative.
It is more professional to use the newton formula (1+a)^m= 1+m*a+m*(m-1)/2*(a^2) etc  which is still
valid for non integral valus of m
( m=.25 in this case) but the calculations are more awkward.
I hope I have been clear enough other wise call back

#34 Re: Help Me ! » Need formula for determining Column and Row » 2012-03-07 21:24:06

I would rather say that
Row= TileNumber div NumberofColumns + 1

#35 Re: Help Me ! » Probability question help » 2012-03-06 21:35:44

Hi bobbym
where is the edit button I can use to edit a post.

#36 Re: Exercises » Bafflers? » 2012-03-06 08:44:13

Results given in post 125 are wrong because i assumed that the replacement operation was carried 9 times. A new simulation assuming the replacement carried 10 times gives an aveerage of 2.812 and 50*(0.75)^10=2.8157

#37 Re: Exercises » Bafflers? » 2012-03-06 08:04:05

The average value calculated from the table (post 125) is 3.76
We could also say that the number of white marbles is multiplied by .75  at each operation so the final result is :
50*(0.75)^9 =3.7541, This method does not seem very rigorous, but the result is practically identical to the simulation result.

#38 Re: Exercises » Bafflers? » 2012-03-06 05:26:48

i made 10000 simulations on computer
The number of white marbles remaining varies from 0 to 13
the distribution is shown below
0             133
1             736
2             1650
3             2120
4             2155
5             1633
6             897
7             437
8             176
9             50
10            9
11            2
12            1
13            1
14            0
15            0
16            0
17            0
18            0
19            0
20            0

#39 Re: Help Me ! » Probability question help » 2012-03-05 09:05:47

I should correct the x table
x=0 if M-M probability 6/15
x=1 if M-F or F-M probability 4/15+4/15=8/15
x=2 if F-F probability 1/15

#40 Re: Help Me ! » Probability question help » 2012-03-05 09:01:15

There were still typing errors. Read
M-M    2/3*3/5=6/15
M-F     2/3*2/5=4/15
F-F      1/3*1/5=1/15
F-M     1/3*4/5=4/15

#41 Re: Help Me ! » Probability question help » 2012-03-05 08:49:53

I am sorry, my previous table is wrong, it should be
M-M    1/3*3/5=6/15
M-F     1/3*2/5=4/15
F-F      1/3*1/5=1/15
F-M     1/3*4/5=4/15

#42 Re: Help Me ! » Probability question help » 2012-03-05 08:38:24

There ae two basic rules in probability theory. If 2 events A and  B with respective probabilities p and q are independent, the probability hat both happen in any order is the product pq. If they are not independent, the probability of A followed by B is the product of p by the probability of B, taking into account that A has occurred.

In the first cases, you have 2 independent events, Success (S in short) with probability 0.98 and Failure (F) with probability 002. After 2 tests there are 4 possibilities
S-S probabllity 0.98*0.98=0.9604
S-F                  0.98*0.02=0.0196
F-S                  0.02*0.98=0.0196
F-F                  0.02*0.02=0.0004
Check that the sum is 1
In the second part events are Male (M)  with probability 2/3 or Female with probability 1/3
if the first is a male ther remain 3 males ad 2 females. Now the probability of M is 3/5 and the probability of a female is 2/5. There fore the probability of M-M is 2/3*3/5=2/5.
Applying this youfind the following
M-M      2/3-3/5=2/5
M-F       2/3*2/5=4/15
F-F        1/3*1/4=1/12
F-m       1/3*3/4=1/4
Check again that the sum is 1

probabilities of the varaible x
probability of x=0  :2/5
      "             x=1  :4/15+1/4=31/60
      "             x=2  : 1/12
Total = 1

#43 Re: Help Me ! » Geometry Problem » 2012-03-05 04:48:07

assuming a circle, the following calculationss made by hand or used as Liberty basic code will give the answer 0.43937252. Sorry, I am not yet familiar with the way to add a figure, but it is not difficult to un derstand what means each line
code
a1=atn((1.53/2)/0.35)
a2=(3.141593-2*a1)
h=sqr(0.35^2+(1.53/2)^2)
d=h/cos(a1)
r=d/2
sat=a2*r^2
st=(r-0.35)*1.53/2
sa=sat-st
answer=1.53*0.53-sa
print answer

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