Two Plus Two Publishing LLC Two Plus Two Publishing LLC
 

Go Back   Two Plus Two Poker Forums > >

Notices

Science, Math, and Philosophy Discussions regarding science, math, and/or philosophy.

Reply
 
Thread Tools Display Modes
Old 09-17-2009, 04:08 PM   #51
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Contradiction.

Suppose m > L. When your sequence gets "close enough" to L, it must also be below m. Choose an epsilon based on m and L that defines "close enough".
Wyman is offline   Reply With Quote
Old 09-17-2009, 11:14 PM   #52
3kingme3
old hand
 
3kingme3's Avatar
 
Join Date: Mar 2008
Location: dont google me
Posts: 1,482
Re: The Official Math/Physics/Whatever Homework questions thread

ok idk wtf Im doing wrong but can certainly use some help.
ln(x) + ln(x - 1) = 1 I have to find x, heres what i did
ln(x(x-1))=1
e^1=x(x-1)
x=e x-1=e so x=e+1
I plugged both back into the original equation and neither worked, but when I submitted no solution I was told it was wrong help anyone?
3kingme3 is offline   Reply With Quote
Old 09-17-2009, 11:27 PM   #53
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by 3kingme3 View Post
ok idk wtf Im doing wrong but can certainly use some help.
ln(x) + ln(x - 1) = 1 I have to find x, heres what i did
ln(x(x-1))=1
e^1=x(x-1)
x=e x-1=e so x=e+1 <------------------
I plugged both back into the original equation and neither worked, but when I submitted no solution I was told it was wrong help anyone?
The arrow is where you went wrong.

If you had 0 = XY, then you can say X=0 OR Y=0. But if you have e = XY, it's not the case that necessarily X=e or Y=e.

So, starting with e^1=x(x-1), get

0 = x^2 - x - e.

Use the quadratic formula to solve for x.
Wyman is offline   Reply With Quote
Old 09-18-2009, 12:08 AM   #54
3kingme3
old hand
 
3kingme3's Avatar
 
Join Date: Mar 2008
Location: dont google me
Posts: 1,482
Re: The Official Math/Physics/Whatever Homework questions thread

tyty so much, apparently I have some shifting problems as well
for some reason I cant get (1/e)^x to shift 10 units right, or
-(1/2)x radical(2x-x^2) to shift 3 units right
Ive tried the obvious subtracting 10 for the (1/e) part and adding 3 inside the radical, but neither have worked
3kingme3 is offline   Reply With Quote
Old 09-18-2009, 12:18 AM   #55
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by 3kingme3 View Post
tyty so much, apparently I have some shifting problems as well
for some reason I cant get (1/e)^x to shift 10 units right, or
-(1/2)x radical(2x-x^2) to shift 3 units right
Ive tried the obvious subtracting 10 for the (1/e) part and adding 3 inside the radical, but neither have worked
To shift right, you replace every (x) with (x-a), where a is the amount of the shift.

Think about why this is the right thing to do.
Wyman is offline   Reply With Quote
Old 09-18-2009, 12:32 AM   #56
3kingme3
old hand
 
3kingme3's Avatar
 
Join Date: Mar 2008
Location: dont google me
Posts: 1,482
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by Wyman View Post
To shift right, you replace every (x) with (x-a), where a is the amount of the shift.

Think about why this is the right thing to do.
Ty again. I would say the reason is because since we are shifting with respect to x's, you need to account for each x in the problem. I guess I got lazy and was thinking about when moving up or down, since you are in respect to y, and there is only 1 y, you just tack it onto the end.

Also I would like to thank you for not just giving me the answer, but rather explaining it to me so I can figure it out.
3kingme3 is offline   Reply With Quote
Old 09-20-2009, 12:26 PM   #57
Subfallen
Carpal \'Tunnel
 
Subfallen's Avatar
 
Join Date: Sep 2004
Location: farther back
Posts: 7,229
Re: The Official Math/Physics/Whatever Homework questions thread

Just to be sure I'm not missing something huge...if A and B are subsets of ℝ2, where:

A = {(x, y) | y < x2}
B = {(x, y) | y x2}

then A and B have the same interior, exterior, and boundary; yes?
Subfallen is offline   Reply With Quote
Old 09-20-2009, 12:39 PM   #58
lastcardcharlie
Carpal \'Tunnel
 
lastcardcharlie's Avatar
 
Join Date: Aug 2006
Location: QED, I think
Posts: 7,335
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by Subfallen View Post
...then A and B have the same interior, exterior, and boundary; yes?
A = int(B) and B = cl(A) so they have the same interior and boundary. Not sure what exterior means, but if it's interior of complement then yes.
lastcardcharlie is offline   Reply With Quote
Old 09-20-2009, 12:58 PM   #59
Subfallen
Carpal \'Tunnel
 
Subfallen's Avatar
 
Join Date: Sep 2004
Location: farther back
Posts: 7,229
Re: The Official Math/Physics/Whatever Homework questions thread

Yeah, Ext A = Int Ac. Thx!
Subfallen is offline   Reply With Quote
Old 09-21-2009, 12:58 AM   #60
furyshade
Carpal \'Tunnel
 
furyshade's Avatar
 
Join Date: Apr 2006
Location: Pasadena, CA
Posts: 10,138
Re: The Official Math/Physics/Whatever Homework questions thread

quick set theory question, is U{{a,b,c,d,e,f},{e,f}} equal to {a,b,c,d,e,f} or {a,b,c,d,e,f,e,f}?
furyshade is offline   Reply With Quote
Old 09-21-2009, 01:31 AM   #61
bigpooch
veteran
 
bigpooch's Avatar
 
Join Date: Sep 2003
Location: Hong Kong
Posts: 3,208
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by furyshade View Post
quick set theory question, is U{{a,b,c,d,e,f},{e,f}} equal to {a,b,c,d,e,f} or {a,b,c,d,e,f,e,f}?
.
bigpooch is offline   Reply With Quote
Old 09-21-2009, 01:39 AM   #62
furyshade
Carpal \'Tunnel
 
furyshade's Avatar
 
Join Date: Apr 2006
Location: Pasadena, CA
Posts: 10,138
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by bigpooch View Post
.
that's what i thought, thanks. also i have a problem involving showing if a set is transitive, "a set A is transitive if each element of A is also a subset of A". maybe i am missing something but i don't see how a set could not be transitive. could someone help me out and maybe give an example of a non-transitive finite set?
furyshade is offline   Reply With Quote
Old 09-21-2009, 02:04 AM   #63
bigpooch
veteran
 
bigpooch's Avatar
 
Join Date: Sep 2003
Location: Hong Kong
Posts: 3,208
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by furyshade View Post
that's what i thought, thanks. also i have a problem involving showing if a set is transitive, "a set A is transitive if each element of A is also a subset of A". maybe i am missing something but i don't see how a set could not be transitive. could someone help me out and maybe give an example of a non-transitive finite set?
Let 0 = {}; then, 0 is "vacously" a transitive set. { {} } = {0} is also a transitive set since 0 = {} is a subset of any set. One can define 1 as {0} and define 2 as {0,1} = { {}, { {} } }, etc.

On the other hand, {1} = { {0} } = { { {} } } is not transitive since 1 = { {} }, the only element, is not a subset of {1}; the only subsets of {1} are {1} and 0 = {}.

There are a lot of "mundane" examples too. For example, any nonempty set S of objects that are not sets will do since each object (by virtue of not being a set) can not be a subset of S.
bigpooch is offline   Reply With Quote
Old 09-21-2009, 02:12 AM   #64
furyshade
Carpal \'Tunnel
 
furyshade's Avatar
 
Join Date: Apr 2006
Location: Pasadena, CA
Posts: 10,138
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by bigpooch View Post
Let 0 = {}; then, 0 is "vacously" a transitive set. { {} } = {0} is also a transitive set since 0 = {} is a subset of any set. One can define 1 as {0} and define 2 as {0,1} = { {}, { {} } }, etc.

On the other hand, {1} = { {0} } = { { {} } } is not transitive since 1 = { {} }, the only element, is not a subset of {1}; the only subsets of {1} are {1} and 0 = {}.

There are a lot of "mundane" examples too. For example, any nonempty set S of objects that are not sets will do since each object (by virtue of not being a set) can not be a subset of S.
so would the the set of natural numbers be a transitive set?
furyshade is offline   Reply With Quote
Old 09-21-2009, 02:33 AM   #65
bigpooch
veteran
 
bigpooch's Avatar
 
Join Date: Sep 2003
Location: Hong Kong
Posts: 3,208
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by furyshade View Post
so would the the set of natural numbers be a transitive set?
If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.
bigpooch is offline   Reply With Quote
Old 09-21-2009, 02:37 AM   #66
furyshade
Carpal \'Tunnel
 
furyshade's Avatar
 
Join Date: Apr 2006
Location: Pasadena, CA
Posts: 10,138
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by bigpooch View Post
If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.
alright, that makes sense. the questions says to show that {null,{null}} is transitive and then asks for an example of an infinitely large transitive set so i figured that is what they were getting at. thanks a lot for the help!
furyshade is offline   Reply With Quote
Old 09-21-2009, 04:04 AM   #67
thylacine
Pooh-Bah
 
Join Date: Jul 2003
Posts: 4,029
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by bigpooch View Post
If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.
Are you sure?
thylacine is offline   Reply With Quote
Old 09-21-2009, 01:36 PM   #68
smcdonn2
centurion
 
Join Date: Jul 2009
Posts: 165
Re: The Official Math/Physics/Whatever Homework questions thread

If probability of having a boy is 1/2. then what is the probability of having three children of the same sex?

I think it is:

1-2[(4 choose 1)(1/2)^1(1/2)^3]

I multiplied it twice becaue it can come 3 boys 1 girl or 3 girls 1 boy

Is this right?
smcdonn2 is offline   Reply With Quote
Old 09-21-2009, 02:25 PM   #69
smcdonn2
centurion
 
Join Date: Jul 2009
Posts: 165
Re: The Official Math/Physics/Whatever Homework questions thread

Lost on this one, Just a move in the right direction would be great

Suppose a series of n independent trials can end in one of three posibilities. Let k_1 and k_2 denote the number of trials that result in outcomes 1 and 2 respectively. Let p_1 and p_2 denote the probabilities associated with outcomes k_1 and k_2. Use theorem 3.2.1 to deduce a formula for the probability of getting k_1 and k_2 occurences of outcomes 1 and 2 respectively.

Theorem 3.2.1 Binomial Distribution

(n choose k)(p)^k(1-p)^n-k
smcdonn2 is offline   Reply With Quote
Old 09-21-2009, 03:41 PM   #70
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by smcdonn2 View Post
If probability of having a boy is 1/2. then what is the probability of having three children of the same sex?

I think it is:

1-2[(4 choose 1)(1/2)^1(1/2)^3]

I multiplied it twice becaue it can come 3 boys 1 girl or 3 girls 1 boy

Is this right?
Clarify: You're assuming that there are 4 children, and you want to know the probability that exactly 1 child is of one sex and 3 are of the other?

If so, p = (2C1) * (4C1) * (1/2)^4 = 1/2.

Which is also what you got, but idk why you subtracted from 1.
Wyman is offline   Reply With Quote
Old 09-21-2009, 03:45 PM   #71
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by smcdonn2 View Post
Lost on this one, Just a move in the right direction would be great

Suppose a series of n independent trials can end in one of three posibilities. Let k_1 and k_2 denote the number of trials that result in outcomes 1 and 2 respectively. Let p_1 and p_2 denote the probabilities associated with outcomes k_1 and k_2. Use theorem 3.2.1 to deduce a formula for the probability of getting k_1 and k_2 occurences of outcomes 1 and 2 respectively.

Theorem 3.2.1 Binomial Distribution

(n choose k)(p)^k(1-p)^n-k
You've got N trials, k_1 of which need outcome 1: N choose k_1
Now you've got N-k_1 remaining, k_2 of which need outcome 2: (N-k_1) choose k_2

The probability of getting k_1 outcome 1's, then k_2 outcome 2's, then outcome 3's is: (p_1)^(k_1) * (p_2)^(k_2) * (1-p_1-p_2)^(N-k_1-k_2)

So your probability is:
(N choose k_1) * ((N-k_1) choose k_2) * (p_1)^(k_1) * (p_2)^(k_2) * (1-p_1-p_2)^(N-k_1-k_2)
Wyman is offline   Reply With Quote
Old 09-21-2009, 03:47 PM   #72
smcdonn2
centurion
 
Join Date: Jul 2009
Posts: 165
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by Wyman View Post
Clarify: You're assuming that there are 4 children, and you want to know the probability that exactly 1 child is of one sex and 3 are of the other?

If so, p = (2C1) * (4C1) * (1/2)^4 = 1/2.

Which is also what you got, but idk why you subtracted from 1.
Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.
smcdonn2 is offline   Reply With Quote
Old 09-21-2009, 03:50 PM   #73
smcdonn2
centurion
 
Join Date: Jul 2009
Posts: 165
Re: The Official Math/Physics/Whatever Homework questions thread

a college awards 5 scholarships
there are 8 men and 10 woman all equally likely to win, what is the probabilty that there will be both men and women in the award?

I did it like this but got a difference answer than the book

(8C0)(10C5)/(18C5)=2.94

1-2.94=97.06%

the book says 96.4
smcdonn2 is offline   Reply With Quote
Old 09-21-2009, 04:15 PM   #74
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by smcdonn2 View Post
Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.
This makes no sense, and you got lucky because 1/2 = 1-1/2.
Wyman is offline   Reply With Quote
Old 09-21-2009, 04:16 PM   #75
Wyman
Carpal \'Tunnel
 
Wyman's Avatar
 
Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 11,955
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
Originally Posted by smcdonn2 View Post
a college awards 5 scholarships
there are 8 men and 10 woman all equally likely to win, what is the probabilty that there will be both men and women in the award?

I did it like this but got a difference answer than the book

(8C0)(10C5)/(18C5)=2.94

1-2.94=97.06%

the book says 96.4
p(all men) = 8C5 / 18C5

p(all women) = 10C5 / 18C5

p(both men and women) = 1 - p(all men) - p(all women)
Wyman is offline   Reply With Quote

Reply
      

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off


Forum Jump


All times are GMT -4. The time now is 06:57 PM.


Powered by vBulletin®
Copyright ©2000 - 2017, Jelsoft Enterprises Ltd.
Search Engine Optimisation provided by DragonByte SEO v2.0.33 (Pro) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.
Copyright ę 2008-2010, Two Plus Two Interactive
 
 
Poker Players - Streaming Live Online