Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Seriously - 100% probability, that somewhere in California or Florida, there are 2 miles of road with 3 or fewer potholes. LOL

"For a given length of highway of 2 miles", with an average of 3 potholes per mile, the probability of 3 or less potholes, is ZERO.

Wrong, the answer is close to zero, but it is not zero itself....
Try Again!!!

Best Regards,
Luke.

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- Luke.

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Seriously - 100% probability, that somewhere in California or Florida, there are 2 miles of road with 3 or fewer potholes. LOL

"For a given length of highway of 2 miles", with an average of 3 potholes per mile, the probability of 3 or less potholes, is ZERO.

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That's good to know, Sarge! Hope to see you back here soon!...........

Best Regards,
Luke.

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- Luke.

Your thread's not been forgotten, Luke.
Myself, I've been incredibly busy. Perhaps after 18 hours of well-deserved sleep, I can solve some of these.

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Capitalize on this good fortune, one word can bring you round ... changes.

Q35, none, new holes always appear faster than they repair them.

lol.... but seriously.... anyone got any answers for these three?.....

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Q36 - Calculus required. There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?

Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

Best Regards,
Luke.

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- Luke.

Q35, none, new holes always appear faster than they repair them.

Bill, you may take a break whenever you like!!!

Also, congratulations to guido.man for successfully completing Q37!!!:

Q37 Answer:

1,000,000 is 11333311 in base 7
so dividing
1 gets $823543
1 gets $117649
3 get $16807 each
3 get $2401 each
3 get $343 each
3 get $49 each
1 gets $7
1 gets $1

Standings Table:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 3 1/2 Points
7. Guido.Man - 2 Point
8. John McLeod VII - 1 Point
9. Labbie - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

With Q37 solved, Guido.man moves up to 7th from 9th...

And then there were 3....

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Q36 - Calculus required. There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?

Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

Have a go & Best Regards,
Luke.

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- Luke.

Daddio will retire for a short while and let others have a whack at these. I do think, though, that these new problems are incompletely specified or require too many assumptions . Of course that's what I thought about the dog and sheep problem before I dug into it.

So Luke may get a lot of questions on interpretations or clarifications.

I will watch the Thread to see the approaches. Working on these has taught a lot about forgotten and new math ideas and how to think "outside the box". It also pointed out tools for computation that exist on the webb.

Here they are, 4 more maths questions....

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Q36 - Calculus required. There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?

Q37 - A man has $1,000,000 he wishes to divide up in his will. He wants to give each person named in his will an amount of money, in dollars, which is a power of 7 ($70=$1, $71=$7, $72=$49, $73=$343, ...). He does not want to give more than six people the same amount. How can he divide the money?

Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

That should keep everyone happy for a while.....

Best Regards,
Luke.

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- Luke.

Well Done Fred!!! for taking out Q32! Bill, you were 2 posts behind!
Q32 Answer: 1 Acre

'Here we are again' Standings Table:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 3 1/2 Points
7. John McLeod VII - 1 Point
8. Labbie - 1 Point
9. Guido.Man - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

Q35, 36, 37, 38 coming soon!

Best Regards,
Luke.

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- Luke.

There are a number of shapes that can be rested against the sides of right triangles and their areas will behave in the x^2 + y^2 = z^2 fashion. Off the top of my head, I think they have to be similar figures.

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Capitalize on this good fortune, one word can bring you round ... changes.

Look at Fred's diagram

The green areas are exactly equal to the red area by pythagoras' theorem or you can compute the areas of each by 1/2 pi (x/2)^2 etc and you will find out that they are equal. But there is one acre of the dog's half circle that does not impinge on the half circles (green) of the sheep.

Therefore when you subtract this from the equality you are left with the dogs having one acre between the two of them. deceptively easy when you convince yourself it can't be hard.

Hope I'm right

I hope you are right, too, as that is what I said 3 hours ago ;))

F.

Yes -I see-- I didn't see Lukes post on this so I assumed that it was still open --Next time I will read all of the posts since some of them contain clues any way. You will get the point I am quite sure.

Before I tried to do an analytical approach; since I was convinced (wrongly) that there was not enough info, I put a sheet of quadrille paper in my printer and printed out a big image of your diagram and hoped it was to scale. I then counted the squares along the base and height and came up with the number of squares for the one acre. I was then going to cut out the extra green areas and then either count those squares or figure out a way to create an accurate balance--But then I had an epiphany and remembered the Indian proof of the Pythagoran theorem and realized the half circles added up just like the sum of the squares of the sides to be equal.

It was a fun problem,

Regards,

Bill

Look at Fred's diagram

The green areas are exactly equal to the red area by pythagoras' theorem or you can compute the areas of each by 1/2 pi (x/2)^2 etc and you will find out that they are equal. But there is one acre of the dog's half circle that does not impinge on the half circles (green) of the sheep.

Therefore when you subtract this from the equality you are left with the dogs having one acre between the two of them. deceptively easy when you convince yourself it can't be hard.

Hope I'm right

I hope you are right, too, as that is what I said 3 hours ago ;))

F.

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Look at Fred's diagram

The green areas are exactly equal to the red area by pythagoras' theorem or you can compute the areas of each by 1/2 pi (x/2)^2 etc and you will find out that they are equal. But there is one acre of the dog's half circle that does not impinge on the half circles (green) of the sheep.

Therefore when you subtract this from the equality you are left with the dogs having one acre between the two of them. deceptively easy when you convince yourself it can't be hard.

Hope I'm right

answer to the sheep problem is one acre total free space outside the fence for the sheep.

Explanation to follow

What do you mean by "outside of the field"? Is the right triangle not completely within the 1 acre field?

Yes, but the rope can twist all the way round the post (360 deg) and the hedge isn't very good. See my picture for how I understood the problem.

F.

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What do you mean by "outside of the field"? Is the right triangle not completely within the 1 acre field?

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Capitalize on this good fortune, one word can bring you round ... changes.

Bill wins yet again!!! :-)

Correct Answer to Q33: 2519
Correct Answer to Q34: Yes
2 more points...

Standings:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 2 1/2 Points
7. John McLeod VII - 1 Point
8. Labbie - 1 Point
9. Guido.Man - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

Congratulations, Bill is now the 3rd Problem Solving Champion!
Also - Sorry, Bill, Fred, etc. you all have Q32 wrong. I'm looking for a answer in acres please...

Q32 - There is a one acre field in the shape of a right triangle, with sides of length x and y. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Tethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?

Best Regards,
Luke.

Message to self: "Read the b******* question!!"

1 acre.

F.

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Bill wins yet again!!! :-)

Correct Answer to Q33: 2519
Correct Answer to Q34: Yes
2 more points...

Standings:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 2 1/2 Points
7. John McLeod VII - 1 Point
8. Labbie - 1 Point
9. Guido.Man - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

Congratulations, Bill is now the 3rd Problem Solving Champion!
Also - Sorry, Bill, Fred, etc. you all have Q32 wrong. I'm looking for a answer in acres please...

Q32 - There is a one acre field in the shape of a right triangle, with sides of length x and y. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Tethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?

Best Regards,
Luke.

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- Luke.

Did I inspire the inclusion of the pictures?
What program did you use to make that, Fred?

Would you believe MS Word copy/pasted into Paint where I added the numbers and the infill colours.

F.

Cool Fred !

How did you generate the BB code ?? Did you start with a JPEG

From Paint, I saved as a .png. Uploaded that to ImageShack (free hosting service). When you upload images to ImageShack, they provide the links for Message Boards, Direct Links, etc.

F.

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