OK... out of Fred's ability to poke fun at the smallest fixture or error I post ;-) I award him a half point! Well Done (and don't do it again! :D)

Onto a more serious matter...
Q13 is won by Jason Gee with the answer: 1/101. Well Done! 1 Point....

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. TBD...

Question 14 (Which I leave with you for the next 9 hours) 1/2 Point:
(23562*24/5)+(732*732/365.52)=

Luke.

Q14: 114563.5225

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For Jason: Advice from an old electrical engineer.


Remember,

Twinkle twinkle little star
E is equal to IR


Regards to the thread,

Bill Rothamel

Twinkle twinkle little star
E is equal to IR

Oh I use that one, I've been using "ELI the ICE man" quite a lot, now I need some rhymes for Norton's and Thevenin's theorems :S

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

OK... out of Fred's ability to poke fun at the smallest fixture or error I post ;-) I award him a half point! Well Done (and don't do it again! :D)

Onto a more serious matter...
Q13 is won by Jason Gee with the answer: 1/101. Well Done! 1 Point....

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. TBD...

Question 14 (Which I leave with you for the next 9 hours) 1/2 Point:
(23562*24/5)+(732*732/365.52)=

Luke.

Q14: 114563.5225

Well Done Fred! Another correct answer!

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 1/2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. TBD...

Now for a one pointer...

Q14: Integral calculus and probability required. Assume that a pregnant woman's probability of giving birth to a girl is p, where p is determined at the mother's birth according to a uniform distribution from 0.4 to 0.6 . If the woman's first child is a girl what is the probability her next child will also be a girl?

Luke.

---

- Luke.

For Jason: Advice from an old electrical engineer.


Remember,

Twinkle twinkle little star
E is equal to IR


Regards to the thread,

Bill Rothamel

And the old Ohm's law pie chart....


Far below Jason's level I am sure....but it always helped me when tweaking electronic circuits.........

---

"Freedom is just Chaos, with better lighting." Alan Dean Foster

For Jason: Advice from an old electrical engineer.


Remember,

Twinkle twinkle little star
E is equal to IR


Regards to the thread,

Bill Rothamel

And the old Ohm's law pie chart....


Far below Jason's level I am sure....but it always helped me when tweaking electronic circuits.........

Jeez - that's a lesson in how to make learning more complicated.

F.

---

For Jason: Advice from an old electrical engineer.


Remember,

Twinkle twinkle little star
E is equal to IR


Regards to the thread,

Bill Rothamel

And the old Ohm's law pie chart....


Far below Jason's level I am sure....but it always helped me when tweaking electronic circuits.........

Jeez - that's a lesson in how to make learning more complicated.

F.

I=E sqaured R.........what's so hard about that? LOL.

As Tom Lehrer once said........the idea is the important thing.......
It's more important to know what you are doing.........rather than to get the right answer.
For research.......here is the clip...Tom Lehrer spoof.......

And a version which has the intro patter........New Math

---

"Freedom is just Chaos, with better lighting." Alan Dean Foster

Now for a one pointer...

Q14: Integral calculus and probability required. Assume that a pregnant woman's probability of giving birth to a girl is p, where p is determined at the mother's birth according to a uniform distribution from 0.4 to 0.6 . If the woman's first child is a girl what is the probability her next child will also be a girl?

Luke.

Well seeing that probability (p - as defined above) is required it would lead to this:
The probability of something occuring (let's call it y) given that a previous event has taken place (z) is the probability of y and z occuring again divided by the probability that z happened. This would mean P(y/z)=P(yANDz)/P(z)
Doing all the maths the answer would be:
38/75 or ~.5066 recurring

---

http://www.therageclub.com

... Q14: Integral calculus and probability required...

Whoops, I'm out, Bit rusty with the old integrals.

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Off-topic but somehow worked its way into this thread,
If anyone out there is familiar with, in electronics, using Norton theorem for DC circuit analysis, I have a circuit comprising a bunch of resistors, a voltage source and a constant current source that I'm wrestling with.

Simulated in Electronics Workbench, and following the prescribed steps, I get the right answer (matches back of book)... but the pencil and paper solution is evading me.

Anyone else familiar with that theorem and perhaps got "Floyd's - Principles of Electric Circuits, Seventh edition'[Chapter 8, Q21 page 340, which relates to a circuit on the previous page] ? (or I could probably email the problem/circuit to anyone that knows that stuff. It isn't an assignment question, just a practice one marked 'extra difficult'.

Jason

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Off-topic but somehow worked its way into this thread,
If anyone out there is familiar with, in electronics, using Norton theorem for DC circuit analysis, I have a circuit comprising a bunch of resistors, a voltage source and a constant current source that I'm wrestling with.

Simulated in Electronics Workbench, and following the prescribed steps, I get the right answer (matches back of book)... but the pencil and paper solution is evading me.

Anyone else familiar with that theorem and perhaps got "Floyd's - Principles of Electric Circuits, Seventh edition'[Chapter 8, Q21 page 340, which relates to a circuit on the previous page] ? (or I could probably email the problem/circuit to anyone that knows that stuff. It isn't an assignment question, just a practice one marked 'extra difficult'.

Jason

Now I'm rusty on that! Been about 20 years since I learned the theorem! Something to do with simplifying circuits using a single power source and risistors in parallel.... if I rememeber correctly!

---

http://www.therageclub.com

Now I'm rusty on that! Been about 20 years since I learned the theorem! Something to do with simplifying circuits using a single power source and risistors in parallel.... if I rememeber correctly!

Yep that's right, figured it out thanks. It was two sources in a weird arrangement (1 voltage source, 1 constant current) that threw me. This one had a resistor in a weird place too, that gets shorted out when making the conversion of one of the sources, so my calculations were out by 2 volts. All good now :D

Jason

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

...Q14: Integral calculus and probability required. Assume that a pregnant woman's probability of giving birth to a girl is p, where p is determined at the mother's birth according to a uniform distribution from 0.4 to 0.6 . If the woman's first child is a girl what is the probability her next child will also be a girl?

Luke.

Well had a look at this again... Still stumped ;D... Anyone?

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Having a look through all the posts, I see BeefDog won it, well done BeefDog!

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 1/2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. BeefDog - 1 Point
7. TBD...

Question 15: (1285*253)/1734+3.1415926=?

Luke.

---

- Luke.

Having a look through all the posts, I see BeefDog won it, well done BeefDog!

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 1/2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. BeefDog - 1 Point
7. TBD...

Question 15: (1285*253)/1734+3.1415926=?

Luke.

Q15: 190.630058574625

F.

---

Having a look through all the posts, I see BeefDog won it, well done BeefDog!

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 1/2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. BeefDog - 1 Point
7. TBD...

Question 15: (1285*253)/1734+3.1415926=?

Luke.

Q15: 190.630058574625

F.

Fred gets that Half-point... He's now gaining in on Jason!

Can You Read This?
1. Jason Gee - 5 Points
2. Fred - 3 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. BeefDog - 1 Point
7. TBD...

Question 16: A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?

Luke.

---

- Luke.

Question 16: A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?

Luke.

Q16: 0.5

F.

---

Having a look through all the posts, I see BeefDog won it, well done BeefDog!

Standings:
1. Jason Gee - 5 Points
2. Fred - 2 1/2 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. BeefDog - 1 Point
7. TBD...

Question 15: (1285*253)/1734+3.1415926=?

Luke.

Q15: 190.630058574625

F.

Fred gets that Half-point... He's now gaining in on Jason!

Can You Read This?
1. Jason Gee - 5 Points
2. Fred - 3 Points
3. Gas Giant - 1 Point
4. JDWhale - 1 Point
5. Dominique - 1 Point
6. BeefDog - 1 Point
7. TBD...

Question 16: A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?

Luke.

Q16: 0.5

F.

Wrong! This question is gonna cause a lot of outrage...

Luke.

---

- Luke.

Question 16: A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?

Luke.

.66 or 2/3

Question 16: A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?

Luke.

Q16: 0.5

F.

Wrong! This question is gonna cause a lot of outrage...

Luke.

Well the only other solution that I can come up with is:
There are 4 possibilities of what we will see when we look at the coin - 3 of these will be heads and one will be tails. Since we see heads, we eliminate the fourth possibility so the probability of the other side being a head is 0.667.

F.

[edit] Rats!! Beaten to it[/edit]

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