- Joined
- 25 February 2011
- Posts
- 5,689
- Reactions
- 1,233
I'm not sure if I understand what you're saying here keith.
4/3 x pi x (h/2)^3 = pi x h^3 / 6
So your initial answer was definitely correct.
No. My initial assumption was that the formula was simply 4/3 x pi x (h/2)^3. i.e the same as a sphere for your example. Trying an example with a non-zero hole radius soon killed that theory and hence my comment that it was a bit more complex.
The wiki page gives the correct formula of pi x h^3 / 6.
I too was heading in the same direction as cynic (Vol of sphere - vol of cylinder - 2 x vol of end cap) when the wiki page pointed me towards napkins.
I'm not sure if I understand what you're saying here keith.
4/3 x pi x (h/2)^3 = pi x h^3 / 6
So your initial answer was definitely correct.
Sorry - meant to type 4/3 x pi x r ^ 3. In my 1st example r did equal h/2.I'm not sure if I understand what you're saying here keith.
4/3 x pi x (h/2)^3 = pi x h^3 / 6
TAKE a look at the two wheels in the image above. Which one is moving faster?
E & 9. The other two cannot possibly violate the stated rule.
As it could be an optical illusion, the easiest way to check is to look at the image type. It is a JPG, so the images aren't moving at all. It would need to be a GIF or some other non-static image type to actually have the wheels moving.
I treated the board as two identical halves - either 2 triangles or 2 rectangles with a line of squares along the axis.
i) choose a line of symmetry (either horz, vert or a diagonal)
ii) place the 1st checker in the centre on the board.
iii) whenever the opponent places a checker NOT on your chosen line of symmetry, you place yours in the symmetrically opposite position.
iv) whenever the opponent places a checker on the line of symmetry, you place yours also on the line of symmetry, but the same distance away from the centre as the opponents.
TL;DR;
The board must always have reflective symmetry. After taking up the only square (the centre) that cannot copied, you simply copy the opponent in the opposite half of the board.
I guess you could also use rotational symmetry.
I've just noticed a flaw in my method which would definitely require additional rules (adding to its complexity) in order to ensure success.I haven't worked through Cynic's detailed answer yet, but Keith's answer is the one I know that will ensure a win.
...
They chose the numbers 6 & 7.
On day 1 (when nobody leaves) they realise that neither could have chosen 11-13 because they are all greater than 10 meaning that the other could be determined by subtraction from 13.
After day 2 they know that nobody could have chosen 0-2 as these can only contribute to a sum of 10 now that 11-13 have been eliminated.
After day 3 they know that nobody has chosen 8-10 as this can now only contribute to the sum of 13 since 0-2 have been eliminated.
After day 4 they know that nobody has chosen 3-5 because these can now only contribute to a sum of 10 since 8-10 have been eliminated.
This leaves only the combination 6 & 7.
Congrats. I agree. The additional bit of info that Mary requires (1 or 2 yellows) is sufficient to distinguish between the duplicate 120s. All the others products are unique & don't require the additional info.2,3,4 & 5.
Y B G R Product Sum
1 2 3 4 24 10
1 2 3 5 30 11
1 2 3 6 36 12
1 2 3 7 42 13
1 2 3 8 48 14
.........
1 3 4 6 72 14
1 3 4 7 84 15
1 3 4 8 96 16
1 3 4 9 108 17
1 3 5 6 90 15
1 3 5 7 105 16
[B]1 3 5 8 120 17
1 4 5 6 120 16[/B]
1 4 5 7 140 17
[B]2 3 4 5 120 14[/B]
2 3 4 6 144 15
2 3 4 7 168 16
2 3 4 8 192 17
2 4 5 6 240 172,3,4 & 5.
There cannot be more than 2 yellow marbles without the sum exceeding 17. So it is known that there are either 1 or 2 yellow marbles.
The fact that Mary asked whether there was more than 1 yellow indicates that Mary has noticed that the house number can be produced from valid configurations containing either number of yellow marbles.
Therefore the house number is 120 as this is the only product that occurs for valid configurations containing exactly 1 yellow and also for a valid configuration containing exactly 2.
The fact that Mary instantly knew from the Father's reply indicates that the number of yellow marbles must be 2 (had there been 1 then Mary would have needed further information in order to to decide between the 2 valid configurations containing 1 yellow that produce the number 120).
The only valid configuration for 2 yellow, which has a product corresponding to 120 and a sum less than 18, is the one containing 3 blue, 4 green and 5 red marbles.
Q1. His strategy will make no difference - the ratio will stay the same at 50%. Each row below represents one set of parents, and each column their successive offspring (stopping when a Girl is produced). The ratio of total number of offspring remains at 50%Increasing the Ratio of Women in the Population.
A king (dirty old leech) decided that there should be proportionally more women in his kingdom than the current 50%. He knows that the probability of a girl at birth is exactly 50% and that males and females have on average exactly the same lifespan (in his kingdom anyway), so he assumes this strategy will work.
He orders that couples should strive to have as many children as possible, but only under the following conditions:
1. If they have a female baby, they are to have no more children.
2. If they have a male baby, they are to continue to have children until condition 1 is satisfied.
Question 1: Will that strategy increase, reduce or make no difference to the proportion of women in the population.
Question 2: With only recourse to edicts of the above kind (when to reproduce and not reproduce etc.) and without recourse to unsavoury methods such as infanticide, forced emigration or things similar, what, if any, would be the optimum policy to adopt to achieve an increased proportion of women in the population. (I know this part is going to cause a lot of controversy).
An easy one that can be solved with no math, just some logic.
A plane flies in a straight line from Perth to Sydney, then back in a straight line from Sydney to Perth. It travels with a constant engine speed (constant air speed) and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from Perth to Sydney?
----
Since air speed is constant, should be the same either way.
So that we are all on the same wavelength, a constant engine speed means a constant air speed, that is the speed of the plane relative to the air it is passing through. Ground speed is the speed of the plane relative to the ground, which will be the air speed + or - the wind speed, depending on whether there is a tail or head wind.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?