Rob_G

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Well
Assume that 12 are already balanced on the scale. If I switch any gear for the 13th one the scale would still remain balanced. Therefore the 13th gear and the gear I would replace have the same weight. I can take the same 13th gear and repeat this for the other 11 gears for the same result making all 13 of them the same weight.

Assuming we can't just break the goblet and spill it all...
I get approx 0.606 in which will displace approx 0.908 cubic in of water.
Oops...I had an extra parenthesis somewhere.
So what I really got was a radius of approx 0.643 in and displacement of approx 1.02 cubic inches.

Maybe...
Anywhere that is exactly 1 mile from the South Pole. 1 mile south to the pole 1 mile "east" which doesn't exist then back north to your original spot.
And...
Anywhere a mile north of where the earth has a circumference of exactly 1 mile.

So...
The multiples of 3 are all of the form 3n where n is a natural number. The largest value of n is floor(1000/3) or 333. So we get 3(1), 3(2), 3(3), ..., 3(333). So if we sum them...sum[n=1...333](3n) we get 3* (333*334)/2 or 166,833.
The same follows for 5 1000/5=200. So sum[n=1...200](5n) or 5*(200*201)/2 or 100,500.
Finally we add them together to get 267,333.

Second part
At an approximately 38.2% success rate going for 2 has a 50% success rate. So anything lower isn't worth the risk.

Assumptions: The other team wont score again, and we will score a second touchdown.
Go for two.
Theoretically, by kicking for the extra point I will score two points. Also, theoretically, I will succeed in one out of the two 2point conversions. In both situations I get two more points and tie to go to overtime, but if I succeed on the first 2point conversion, I can kick a field goal after the second touchdown and win.
So doing the numbers:
Extra point: 100% chance of a tie and 50% chance of winning or if you go for one then the second time go for two you still only have a 50% chance of winning. The only difference is what you put your hope for the win in.
Two Points: You have a 50% chance of converting the first one and when you do you have a 100% chance of winning by kicking a field goal.
If you miss the first (50%) you have a 50% chance to make the second so a 25% chance of that outcome. In this case you only tie and go to overtime with a 50% chance of winning or 12.5% overall. Putting this together with the first part gives us a total of 62.5% chance of winning.
The other part will require paper instead of just my head so I will get back to you with that.
 1

What is the current official highest N? 4?

Is that supposed to be median or mean? Because d has no effect on the median.
Unless you meant the median of c and d.

Oops! In that formula
y^2 should be A. I was working stuff out separately and for got to put the sqrt(A) back in.

I'm thinking the length of the radius is .5(x+(y^2)/x).
Again I'm posting from my phone as my laptop needs to charge. I'll show my work when I get a chance to plug it in.

Let's hope I do this right. I'm posting from my phone.
So assuming the same as above bishop on f1 and knight on g1 I would say:
The bishop can reach any square in 1 or 2 moves.
The knight takes at most 5 moves to reach a square as stated above.

Given angle a find ceiling(90/a)=n a*n will be the angle made where the light will start to come backout. The distance from the intersection should be d/sin(n*a)
 1

When the light starts to come back out it is d units from the intersection along the top line. And as long is my thought process is correct the light will follow the exact same path in and out of the mirrors.

If we take before to mean infront of, his father watched his birth, but his mother died giving birth. The man then grew up to be a priest/pastor/justice of the peace/etc and officiated his sister's wedding.

Well I guess it depends on how the store operates. I could justify 3 different answers.
$75 because most stores don't allow you to use coupons on bogo sales.
$67.50 because they take 10% off of the $75 that you would be charged.
$60 because the store first charges you for both, takes the 10% off the takes off the $75 for the bogo sale.

27?
The shape is a square and adding 27 to the total would make is a square number. 784
 1

no.
There are 'technically' 16 possible outcomes some are just equivalent.
1,1 2,2 3,3 4,4 all have a 1/16 chance of occuring while the rest have a 1/8 chance.
1,1 2,2 3,3 4,4
1,2 2,1 2,3 3,2 3,4 4,3
1,3 3,1 2,4 4,2
1,4 4,1

(1Pi/360)^{n2 }
I'm not certain my math is correct though.

The first three points are always within a semicircular arc.
From 4th point onwards, the probability of lying within a semicircular arc is 1/2. For 4th the probability is 1/2. For 5th, it will be 4th lies within the arc and the 5th also; so 1/4... and so on
The overall probability is then 1/2^{(n3)} for n>3. For n<3 it is 1.
with your first statement. A counterexample would be if the three points formed an inscribed equilateral triangle.

There is no error.
We know p/n+q/n = (p+q)/n
In his formula pq can be written as 10p+q, qp as 10q+p and nn as 10n+n.
So (pq+qp)/nn = ([10p+q]+[10q+p])/(10n+n) = (11p+11q)\11n = (11[p+q])/11n = (p+q)/n = p/n+q/n.

37
The missing guess is either 24 or 50.
The missing difference is 8.

there are 3,346,559 ways to write the second code and not repeat a triplet.
because?
10!  ( 8*8!  7*7!  6*6!  5*5!  4*4!  3*3!  2*2!  1*1! )

3: 147 rows of 4
4: 116 rows of 5
5: 95 rows of 6
7: 68 rows of 7
9: 51 rows of 10
11: 39 rows of 12
14: 26 rows of 15
19: 11 rows of 20
23: 2 rows of 24
24: no main block

there are 3,346,559 ways to write the second code and not repeat a triplet.
twins age
in New Logic/Math Puzzles
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