A really simple question.

Hummmm!

Gotta cogitate on this one for a while!

The only thing that seems to make any sense at all as a response to this puzzle is:

The volume is the same as it was before you did the drilling -- and the hole does not impact on the volume of a sphere.

Although I have a BA in Economics and worked on my MA in Psychology (never finished) I truly have never taken any math course above the level of arithmatic. Don't ask how I managed to pull it off -- I'm a crafty mutha. But I mention this because I'm not sure if the volume of a sphere is effected by a hole -- or by what is contained in the sphere. I suspect not!)

Frank

Surely what you are saying is that if you drill a hole in a sphere you have removed nothing from it?

If the sphere is wooden those wood shavings do not exist?

That if you drill a hole that is only marginally smaller in diameter than the sphere, leaving only a circlet of foil, the sphere is still as heavy as it was?

That - intuitively - does not sound a very safe assumption :-)

No, that's not what I'm saying at all.

I'm saying that even if you remove the entire of the inside of a sphere -- the dimensions of the sphere stays the same.

A solid sphere and a completely empty sphere of the same dimensions (or so it seems to me, restricted as I am by what I wrote earlier) -- have the same volume.




Yep! They exist.




Nope, it ain't as heavy. But you asked about volume -- not weight.




Well it didn't earlier when I first proposed it -- but to be honest, it does now.

I think you are just pulling my leg me to see how I react.


By the way, I noticed that you did not say I was wrong.

Any reason for that?

Frank

I did not say you were wrong simply because you were neither wrong nor right. Had you proposed a solution that could have been right or wrong, but what you did was to suggest an approach. The trouble with saying an approach is right or wrong is that only no approach at all can be certain of leading nowhere.

A clue: if you were to stand your reasoning on its head . . .

You said - I think - that volume has little to do with weight. I suggest otherwise. If you fill a tank with liquid it will get heavier - weight - as it gets more full - volume. A cubic meter of lead - volume - is not as light - weight - as a cubic centimetre of lead - volume again. And if you drill a hole in a sphere there is less space that is 'sphere-space' and more that is 'not sphere-space' after the hole is drilled.

If the drilling removes all of the material of the spere than, regardless of what was once 'sphere space' and what could be sphere space, there is no longer any space at all that is occupied by that sphere qua sphere.

Besides all of that, I did ask the volume of 'what remains of the sphere'. I meant that literally and in its everyday meaning.

You are starting to become a lot less fun!

Frank

I'm sorry to hear that. (Not that I ever considered myseelf to be the acme of fun)

The last post I wrote - if that is the problem - was written in a hurry. I wrote it under the impression that I had not explained myself too well and my main aim was to make it all clear. If I was also over-serious and earnest, put that down to trying to do three or more things at once.

But fun? When was I ever fun?

Try

Quantum mechanics? Me? I wouldn't even know which wrench to use. Couldn't fix a quantum to save my life. Gee willikens, boy, tractor mehanics is tough enough for me.

Anyways, nobody in my village has got a quantum so there's no call for no mechanicking.

You would need a very small wrench!
(
4/3) * PI * r^3 where r is the radius of the sphere.

Cards.
x^2 = (x/[exponent])*(x+x), when x is even
x^2 = x * (x+1) - x, when x is odd
576^2= (576/2) *(576+576) = 288 * (1152) = 331776
576^2 = 331776
573^2 = 573 * (573+1) - 573 = 573 * 574 - 573
3238902 - 573 = 328329
573^2 = 328329. Now will it work for smaller numbers?
Now this may seem like more work but it will help to explain the mechanical action behind powers and roots.
Please, if you have any input on how to calculate
x^3 for both even and odd numbers let me know.

If the failure rate is constant over time,
Then R = e^-(lambda x t)
Where R is the probability of success
e is the base of the natural logarithm
lambda is the failure rate
t is the operating time.

Failing that.
E = Mc^2
E = Energy
M = Mass
c = the speed of light (2.99792458e8 meter/sec)
In other words:
Energy = Mass * 89875517873681764
This implies that if mass could be converted to energy in a controlled environment, 1 gram of a mass would produce 89,875,517,873,681,764 units of energy.

I hope that answers any further problems you may post.

Ps. I realize it is easy but, three (3) new students and teacher says she has 210 seating permutations. How many actual seats are empty

Quickie.
The average of 6 numbers is 19. A seventh number is added and the new average is 20. What is the 7th number

6?

Try

And please note, this is not a fun reply.

A small wrench? You do not realise how much I love being a tractor mechanic. I mentioned the possibility of my becoming a quantum mechanic to my buddy, Ephraim, and he said that if I did I would have to move to a big city, somewhere like Jackson Fork, because if anyone anywhere has a quantum that needs a mechanic, it would be Jackson Fork. Give up being a tractor mechanic and move to the Concrete Jungle of Jackson Fork? You just don't know how big a wrench that would be.

(BTW; you may not know of Jackson Fork but believe me it is BIG!! I am told they have three filling stations there and do you realise just how many pick-ups it takes to keep three filling stations in business?)

Yep; you are right about the volume of a sphere and if we ever need to know the volume of a sphere you will be the first one we will call. :-) But we do not know the size of the sphere.

I certainly agree that as the radius of the hole approaches zero the length of the hole approaches the diameter of the sphere, so it would be possible to find an answer using 1.5 units as the radius of the sphere.

But this is by no means shown to be true for any other radius of the hole other than zero and holes with a radius of zero are few and far between - or ubiquitous, depending on how you look at it!

You are kidding me, right? That 'averages' question just needs (20*7) - (19*6) = 26. So where does 'E = MC^2' come into it?

Accuracy, brevity and speed. Frank, you excel at them all. Well, that's not strictly true, but at least two. If 6? Is your answer to the empty seats. Then you were close. If it was for the Quickie, then one number is right.

Mungo, its Jerry, is it not. How I loved those 70's songs.
Ephraim, my word it has been a few years since I was fishing in Eagle Harbour.
Jackson Fork must be over 200 miles south. And I am sorry, but I do not believe it has three trucks. Take my advice try Madison or if you are still set on tractors. Try Kearney and Trecker in Milwaukee.

Quote…" for any other radius of the hole other than zero and holes with a radius of zero are few and far between - or ubiquitous, depending on how you look at it!"
In future, if you are going to come up with such lines, please preface them with a health warning. I laughed so much my nose was sick.

Quote…" But we do not know the size of the sphere." I used to go hunting with a Spear and it was about 8ft long. (I hope that helps)

26 is only just on the right side of correct, and as this is your thread, I will give you the benefit of the doubt. E=MC^2 Do not ask me have a word with Einstein.

Trial

Your 'seats' question does not make it clear if the teacher (t) occupies one (1) of the seats or if three (5) "new" students implies that there are a number (x) of "old" students in addition to those three (00000011) However, the possible answers suggest that we only have to consider three (2π) butts (b) on seats.

That being so, three (√65) students on seven (18¼%) seats would give two hundred and ten ($3.25) combinations. With seven (4!) seats and three (≈2.999) butts then four (Φ) seats would be empty.

What a lucky guess. :wink: The formula for an r-permutation from a set of n elements is n!/(n-r)! Here n is unknown but you know that r is 3, so
n!/(n-3)! = 210
n*(n-1)*(n-2) = 210
n=7 (7*6*5=210)
So there were 7 empty seats before seating the 3 new students.
Is there, I ask, any formula you dont know? :wink:

My answer of 6 was for the Quickie.

But now I am saying "Oops."

I suspect the number actually is 26.

My reasoning:

Here is the problem as stated: The average of 6 numbers is 19. A seventh number is added and the new average is 20. What is the 7th number.

If the average of 6 numbers is 19 we know that the total of the numbers has to be 114 (6 x 19 = 114)

In order for 7 numbers to average 20 -- the total of the numbers has to be 140 (7 x 20 = 140)

The difference is 26 -- which should be the solution.

Is it?

Another "oops."

I just read Mungo's answer while looking over the seating puzzle. I had missed that question.

Apparently I was right at 26.

I did get it on my own -- and I know how I screwed up the first time.

Remember, I gotta do this stuff longhand!

Well it looks to me as if we old tractor mechanics of Dodgetty Creek has got you-all city folks tubularised on that thar Britney Spheres question.

So lookee here:

Volume of Sphere; (4πr^3)/3
Volume of "end caps" - the recipe calls for two of these; πh(3R^2+h^2)
Volume of cylindrical hole; πR^2L

Where 'r' is the radius of the sphere
'R' is the radius of the hole and of the face of the 'end caps'
'h' is the height of the 'end caps'
'L' is the length of the hole.
(and 'π' is the best this tarnation font can do as regards 'pi')

So the remaining volume once the hole has been drilled through the middle has to be:
Sphere volume minus end caps volume minus hole volume.

(4πr^3)/3 - 2πh(3R^2 + h^2) - πR^2L

I'll leave you-all to have a go at doing the working out on that lot but it simplifies to:
πL^3 / 6

The length (L) of the hole was given as three units, so this becomes:

9π/2 ( = 14.137 cubic units to 3dp)

As a check we can consider the limiting case where 'R'= 0 and where both the hole volume and the end caps have zero volume. 'r' would in this case equal 1.5 (half of that three) and the volume of the original sphere should be the same as the previous result.

(4πr^3)/3, (where 'r' = 1.5) is (4π(3.375))/3 = 14.137. Which does tally.

(I did say it was a 'hard mathematics problem' did I not?)

What intrigues me about this question is that so little information is needed.

a) Frank, you are a star.
b) Mungo, Harry, Jerry, you have the same answer as I.
c) But then, I have never been to Dodgetty Creek. Ha!ha!

Mungo

I've thought this over at some length and I think your "answer" to the volume of the sphere poser is incorrect...and that my proposed answer id correct.

First of all, to suppose, as you assert, that "weight" has anything to do with the "volume" of a sphere -- is just plain wrong.

"Volume" in the mathematical sense being used here is defined as: Space occupied in three dimensions, as measured by cubic units; the amount of space included by the surfaces of a solid.

Before proceeding, let me deal with your wording. You wrote: "The question is, what is the volume of what remains of the sphere?" You mentioned that in your reply to my suggestion. But the "volume" of a solid -- in this case a sphere -- has nothing to do with the "solids" that remain inside the sphere. The space inside the sphere -- composed in this instance of both solids and air -- is the volume.

I was absolutely correct in saying that the volume of the sphere in question is the same if it is filled with a solid material or with air or with a combination of solid and air.

The volume of the sphere is 4xr>3/3 whether it is filled with wood, filled with air, or filled with some wood and some air.


If you disagree, let's discuss it.


Try; You seem to be the only other one here. Do you have an opinion on this?