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PostPosted: Tue May 17, 2005 11:59 am 
Way Beyond Godly
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Hunter, you are awesome. Mind if I kidnap you so you can help me with maths all the time? :D

Now.. anyone happen to know the answer to question 1? :battar:


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PostPosted: Tue May 17, 2005 2:35 pm 
PPT Toddler
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1) Two bowls each contain 8 pieces of fruit. In bown A there are five oranges and three apples; in bowl B there is one orange and seven apples.
-a- For each bowl, find the probability that two pieces of fruit chosen at random will both be apples.
-b- For each bowl, find the probability that two pieces of fruit chosen at random will both be apples, when the first piece of fruit is replaced before the second is chosen.
-c- One bowl is chosen at random and from it both pieces of fruit are chosen at random without replacement. If both pieces of fruit are apples, find the probability that A was the chosen bowl.
-d- One bowl is chosen at random, and from it two apples are chosen at random, the first apple being replaced before the second is chosen. If both pieces of fruit are apples, find the probability that bowl A was the bowl chosen.


-A-

The probability of both fruits being apples is equal to the probability of drawing an apple the first time and then doing so again the second time.

Bowl A: 3/8 * 2/7 = 3/28
Bowl B: 7/8 * 6/7 = 3/4

-B-

Same procedure as part A, but because the fruit is being replaced, the chances of drawing an apple the second time is the same as drawing an apple the first time.

Bowl A: 3/8 * 3/8 = 9/64
Bowl B: 7/8 * 7/8 = 49/64

-C-

Think about this one logically for a moment. If Bowl A has a 50% chance of drawing two apples and Bowl B has a 25% chance, what would your answer be? The probability of the bowl being Bowl A would be twice as much, right? That is, a 2/3 chance of the bowl being A, and a 1/3 chance of the bowl being B.

The mathematical form of this logic is:
Chance of Bowl A = Bowl A's chance of drawing two apples / (Bowl A's chance of drawing two apples + Bowl B's chance of drawing two apples).
Chance of Bowl B = 1 - Chance of Bowl A.

Because this is without replacement, we use the figures from part A.

Bowl A: (3/28) / (3/28 + 3/4) = 1/8
Bowl B: 1 - 1/8 = 7/8

-D-

Same idea as part C, except now the fruits are being replaced.

Bowl A: (9/64) / (9/64 + 49/64) = 9/58
Bowl B: 1 - 9/58 = 49/58

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How do you differentiate e to the power of x to the power of (3-x)?
We have not learnt how to deferentiate x with x as the power.


Because e is your base, you don't need to do anything fancy just because you have x as the power. Because it is a composite function, this is simply applying the chain rule.

(e^(3-x))' = e^(3-x) * (3-x)' = -e^(3-x)


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PostPosted: Tue May 17, 2005 3:37 pm 
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What does the symbol ' mean? And what do you mean by not doing anything fancy? I just need to ignore the x?

EDIT: Sorry, but I think you misunderstood what I meant. The x, which is the power of e, has its own power which is (3-x). It's not e^(3-x), more like e^x^(3-x). Oh, and the answer from the answer sheet is something like x²e^[x(3+x)].


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PostPosted: Tue May 17, 2005 4:09 pm 
PPT Toddler
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843 wrote:
What does the symbol ' mean? And what do you mean by not doing anything fancy? I just need to ignore the x?

EDIT: Sorry, but I think you misunderstood what I meant. The x, which is the power of e, has its own power which is (3-x). It's not e^(3-x), more like e^x^(3-x). Oh, and the answer from the answer sheet is something like x²e^[x(3+x)].


The symbol ' means "prime"; f'(x) is pronounced "f prime of x" and it stands for the derivative of f with respect to x.

And yes, I misunderstood your question. To compute this derivative, some nasty calculations are in order.

Let y = e^(x^(3-x))
Take the natural log of both sides to eliminate e.

ln(y) = x^(3-x)
Take the natural log again to allow you to bring down the (3-x).

ln(ln(y)) = (3-x) * ln(x)
Implicitly differentiate with respect to x.

1/(y*ln(y)) * dy/dx = 3/x - ln(x) - 1
Solve for dy/dx by multiplying each side by y*ln(y). Remember, y = e^(x^(3-x)) and ln(y) = x^(3-x).

dy/dx = (3/x - ln(x) - 1)(e^(x^(3-x)) * x^(3-x))
This mess can be marginally simplified to:

-x^(2-x) * (x*ln(x) + x - 3) * e^(x^(3-x))


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PostPosted: Tue May 17, 2005 4:26 pm 
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Wow, thanks a lot. You really are an 'Angel of Mercy' :lol:


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PostPosted: Tue May 17, 2005 7:39 pm 
Way Beyond Godly
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Man...I want to help someone, but all this maths is going over my head.

EDIT - Optimus: if you need any help on your assignment, I could possibly help.


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PostPosted: Sat Jun 04, 2005 5:21 pm 
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Think anyone could help me with this one?

Quote:
When the price of an item was lowered by 20%, its sales (money winnings) went up by 4%. How many percent did the sales of the actual item go up (meaning how many more items did the sell than before)?


Let me know if I need to clarify anything, I traslated this from Finnish so I'll blame it on that however :P

The answer should be 30%.


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PostPosted: Wed Jun 08, 2005 5:47 am 
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Quote:
When the price of an item was lowered by 20%, its sales (money winnings) went up by 4%. How many percent did the sales of the actual item go up (meaning how many more items did the sell than before)?


Easiest way is to plug in numbers. Say the original price was $100. Lowering the price by 20% makes the new price $80.

Now say we originally sold 100 items at $100 apiece, hence $10000. When we lowered the price to $80 we sold 104% of $10000 worth, or $10400. How many items is this? Well...

$10400 / $80 (per item) = 130 items.

So originally we sold 100 items; once we lowered the price, we sold 130 items. The answer: We sold 30 more items, or 30% more than before.


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PostPosted: Thu Jun 09, 2005 3:05 pm 
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Can someone help me with this problem?



A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The voltage across the primary circuit is 7.2 kV. What voltage is across the secondary circuit?



...That's it. I've no idea where to begin and it's the last problem on my homework (why this teacher always makes insane problems last I've no idea).


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PostPosted: Fri Jun 10, 2005 7:44 pm 
Beyond Godly
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kuronue wrote:
Can someone help me with this problem?



A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The voltage across the primary circuit is 7.2 kV. What voltage is across the secondary circuit?



...That's it. I've no idea where to begin and it's the last problem on my homework (why this teacher always makes insane problems last I've no idea).


From what I know you would do

7500/ 125= 60

7.2/ 60= 0.12kV = 120V

I'm not 100% sure about this but that's how I would have worked it out...


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PostPosted: Sun Jun 12, 2005 8:56 pm 
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I have an accounting assignment and I just don't understand this part, can anyone explain this to me?

Quote:
Telephone Bill for December 2004 is received in January 2005, $85.20, Automobile Tune up $400, Photocopying $60.30. Gst Recoverable as of December 31, 2005 is 516.80.

These amounts include GST. Calculate this GST by dividing each amount by 1.15 and multiplying the result (base) by 7%. The Accounts Payable Credit will be this total including the GST Recoverable. Add up the three GST amounts and record a total debit to GST Recoverable in the adjustments column.


For GST Recoverable I got, $33.21 and for Accounts Payable I got $550.01. And the totals at the bottom still don't add up. I don't understand if I'm doing something wrong.


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PostPosted: Tue Jun 14, 2005 6:27 pm 
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Sorry to bother you, but I got a quick question about Conductivity of an electrolyte. Our chemistry teacher is... lacking to say the least, so I need a bit of a hand. ANy help appreciated.

Bascally, we did an experiment which tries to find out the relationship between the distance apart of two electrodes, and the effect that has on the conductivity. We know the distance apart (Independant Variable), we know the voltage (Constant - 6V), we know the current (Dependant Variable), we know the depth at which they are submerged (Constant - 5cm), we know the width of the beaker (Constant - 8cm), we know the depth of the electrolyte (Constant - 5cm), but where does this all fit in into the conductivity formula:

Conductivity = (Current / Voltage) * (Length / Area)

I'm not sure if I got the wrong formula, but I find it hard to equate Length and Area into this formula. Perhaps this will only work in wire...?

I'm baffled :(

Any help is very much appreciated. If not for this homework, then for future experiments.

Once I've done my chemistry, Im sure I'll have some Maths work to stick up here as well. :roll:


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PostPosted: Wed Jun 15, 2005 10:02 am 
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kuronue wrote:
Can someone help me with this problem?



A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The voltage across the primary circuit is 7.2 kV. What voltage is across the secondary circuit?



...That's it. I've no idea where to begin and it's the last problem on my homework (why this teacher always makes insane problems last I've no idea).


Hey! That's from Subeta!


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PostPosted: Sun Jun 19, 2005 3:32 am 
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^_^ I cant table this answer..

A farmer buys 100 live animals for $100, how many of each does he buy if chickens are 50cents each, goats are 3.50 each and sheep are $10 each.

Theres loads of different answers yeh?

Show how a table can be used to solve this question

I hate tables, why couldn't he just buy 10 goddamn sheep and be done with it!


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PostPosted: Sun Jun 19, 2005 4:24 am 
Beyond Godly
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zorg wrote:
^_^ I cant table this answer..

A farmer buys 100 live animals for $100, how many of each does he buy if chickens are 50cents each, goats are 3.50 each and sheep are $10 each.

Theres loads of different answers yeh?

Show how a table can be used to solve this question

I hate tables, why couldn't he just buy 10 goddamn sheep and be done with it!


I got 92 chickens, and 4 goats and sheep. o.o

Sorry if I'm wrong. I did it while laughing at Weekend Update.


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