The Official Math/Physics/Whatever Homework questions thread - Page 3

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".

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09-17-2009 , 11:14 PM

#52

3kingme3

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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?

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09-17-2009 , 11:27 PM

#53

Wyman

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Quote:

Originally Posted by 3kingme3

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.

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09-18-2009 , 12:08 AM

#54

3kingme3

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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

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09-18-2009 , 12:18 AM

#55

Wyman

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Originally Posted by 3kingme3

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.

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09-18-2009 , 12:32 AM

#56

3kingme3

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Join Date: Mar 2008 Posts: 1,483

Quote:

Originally Posted by Wyman

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.

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09-20-2009 , 12:26 PM

#57

Subfallen

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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?

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09-20-2009 , 12:39 PM

#58

lastcardcharlie

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Quote:

Originally Posted by Subfallen

...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.

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09-20-2009 , 12:58 PM

#59

Subfallen

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Yeah, Ext A = Int A^c. Thx!

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09-21-2009 , 12:58 AM

#60

furyshade

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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}?

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09-21-2009 , 01:31 AM

#61

bigpooch

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Quote:

Originally Posted by furyshade

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}?

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09-21-2009 , 01:39 AM

#62

furyshade

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Quote:

Originally Posted by bigpooch

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?

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09-21-2009 , 02:04 AM

#63

bigpooch

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Quote:

Originally Posted by furyshade

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.

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09-21-2009 , 02:12 AM

#64

furyshade

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Quote:

Originally Posted by bigpooch

so would the the set of natural numbers be a transitive set?

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09-21-2009 , 02:33 AM

#65

bigpooch

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Quote:

Originally Posted by furyshade

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.

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09-21-2009 , 02:37 AM

#66

furyshade

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Quote:

Originally Posted by bigpooch

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!

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09-21-2009 , 04:04 AM

#67

thylacine

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Quote:

Originally Posted by bigpooch

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?

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09-21-2009 , 01:36 PM

#68

smcdonn2

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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?

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09-21-2009 , 02:25 PM

#69

smcdonn2

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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

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09-21-2009 , 03:41 PM

#70

Wyman

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Quote:

Originally Posted by smcdonn2

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.

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09-21-2009 , 03:45 PM

#71

Wyman

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Originally Posted by smcdonn2

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)

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09-21-2009 , 03:47 PM

#72

smcdonn2

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Quote:

Originally Posted by Wyman

Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.

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09-21-2009 , 03:50 PM

#73

smcdonn2

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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