dgreening
-
Posts
125 -
Joined
-
Last visited
-
Days Won
13
Content Type
Profiles
Forums
Events
Gallery
Blogs
Posts posted by dgreening
-
-
Instead of simply rounding up or down, consider how much the percentage change would be for each person compared to what they would pay [if coins were available].
consider Charlie
If the bill is even [$10], then he must pay an even number of $$
If the bill is odd [$11] then he must pay an odd number of $$
For a $10 check
The actual amount due from AB and C is [4.29, 4.29, 1.43]
you could charge Chrlie 0, $2 or $4
Leaving the others to pay $5, $4 or $3
[4,4,2] would give A and B a 7% break and C would overpay by 40%
[5,5,0] would have A &B overpay by 17% and C would get a 100% break
Looking at the differences, [4,4,2] yields the lowest absolute % offset of 54% (7 + 7 +40).
[5,5,0] would yield an absolute offset of 112% (6,6,100)
So for a $10 check, the fairest split is $4 from A & B and $2 from C
For $11
The choices are [5,5,1] and [4,4,3]
[5,5,1] over charges A&B by 6% and give C a 36% break
total offset = (6+6+36) = 48%
[4,4,3] gives A&B a 22% reduction and overcharges C by 75%
total offset = (22+22+75) = 119%
So the fairest split is $5 from A and B, $1 from C -
"it's only true if..."
Nice reply! ,,, even if it is a mouthful!
-
Yes, I would take the bet ... but I do not have a strong proof
Just based on odd and even numbers if we get 2 even numbers they are
relatively prime. So 1/4 of the time you would expect a positive pay off. If
for a moment we rule out any odd number as being
co-prime, we would get -1 three times out of fourThe worst expected pay off [based on this poor assumption] would be
P*min = (1*(+2) + 3*(-1))/4 = -1/4
If just 1 out of 9 combinations involving an odd number were relatively
prime, the Pay off would go to zero [making it a fair game].[this is where things get shaky].
I have examined numbers up to 65 and found that if at least one number is
odd, the chances of the pair being relatively prime is closer to 2/9 which would
return a positive expected pay off.I am sure that someone out there has an elegant solution, that I look forward
to seeing! -
Actually
the second time the minute hand was 7 minutes BEHIND the hour hand
first position 1:12
second position 8:36
elapsed time 7:24 [assuming it was in the same 12 hour period]
I guess it helps if you read the OP correctly! DUH
-
Actually
the second time the minute hand was 7 minutes BEHIND the hour hand
first position 1:12
second position 8:36
elapsed time 7:24 [assuming it was in the same 12 hour period]
-
The Best Answer still bothers me.
The OP states that the plane flies "at constant airspeed c directly above a closed polygonal path in a plane".
If the desired path over the ground is [east to ] west at a constant "Air speed" and the wind is blowing due north at some velocity, the place will tend to be thrown off course and it will result in some angle [similar to that shown by K-Man above]. As a result the actual distance moved in the E-W direction will be reduced by the resulting angle.
To compensate for this, a pilot must assume a constant "crab angle" in order to stay above the desired line of flight that is choosing a vector that adjusts for the crosswind. The same trig still applies and the "ground speed" will be reduced.
If the wind direction and speed are constant over the entire course throughout the flight, the ground speed will be reduced similarly on some portion of the return flight.
-
"Composite" refers to the number having at least two distinct positive integer divisors.
Actually, that's not quite true as 1 is also a positive integer divisor.
Composite really just means "not prime".
Thanks
-
Not familiar with the term in this usage - What do you mean by composite??
-
Red Herring!???
Please say it isn't true.It's not possible for n=7. Suppose it was possible and we found 2 coins with probabilities P1 and P2. Using these 2 coins we can generate events with probabilities that can be expressed as one of the following:
- P1nP2m
- (1-P1)nP2m
- P1n(1-P2)m
- (1-P1)n(1-P2)m
Since 7 is prime, the only way to generate probability 1/7 is to have either P1 or P2 equal 1/7. But if one of the coins has probability distribution of 1/7:6/7 then the other coin must allow us to generate 6 equally likely outcomes, which is impossible. Note that the 1/7:6/7 coin is useless after the first flip.
bonanova's example is a red herring.
-
Question
The OP implies that the unfair coin in your example goes to one side 2 out of 3 time.
Can you choose the coing such that you get a specific degree of "unfairness" [e.g., Heads 3 out of 5 times]?
-
Some initial thoughts
About the dice
- There are 168 prime numbers between 1 and 1000 [i looked it up]
- Therefore there are 832 non-primes in the range
- The average of these 168 numbers is about 453.1
- All except for "2" are odd numbers [which will come up about 0.6% of the time]
- The distribution is not close to uniform
- For any range, the primes with lower values, tend to be more represented
- 25 primes are below 100, while only 13 primesare between 900 and 1000
- there are 95 primes in the range 1-500, while there are only 73 primes from
500 to 1000
- For any range, the primes with lower values, tend to be more represented
Since about 99.4% of all rolls will result in an odd number
- If you simply roll the dice and sum the results -
- An even number of rolls will alomst always result in an even number
- An odd number of rolls will almosat always result in an odd number
Based on these observations, you cannot simply roll the dice a specific number of items and add them
- The results will be biased to odd or even
- low numbers [like 4 or 20] will almost never occur.
The algorithm must be more complicated than simple summation.
- There are 168 prime numbers between 1 and 1000 [i looked it up]
-
A mathematician / logician / perfectionist should choose the first.
Being an engineer, I prefer the second.
Here, here!
-
Wouldn't a triangle with one side = 0 collapse into a line segment or 2
co-located line segments ??I don't have an answer, but I suspect that is NOT the answer that BMAD is
looking for - but it made me chuckle [thanks].What immediately comes to mind is triangles with one side length zero. As for the second question, it probably holds for any figure as long as they are similar and the shared side is the same.
-
Yes. There are 2 answers
This occurs at about x= 0.27 pi
It also occurs at about x= 0.93 pi
f(0.27 pi) = f(0.93 pi) = about 12.0
-
Following the earlier post, let's look at both parts of the equation
9*x*(sin x) +4/(x*sin x)
over the range of 0 to pi
9*x*(sin x) starts at 0, ends at 0 and would expect it to peak somewhere past 0.5 pi
4/(x*sin x) starts very large, ends relatively large and you would expect it to reach a minimum somewhere before 0.5 pi
Since the peak of one function and the minimum of the 2nd function do not coincide, You would expect 2 relative minima.
Rearranging 2 pieces, we get to
9 * (xsinx) = 4/(xsinx)
or
(xsinx)^2 = 4/9
therefore
xsinx = sqrt (4/9) = 2/3
I am sure that there is an elegant way to solve this, but I cheated
This occurs at about 0.27 pi
It also occurs at about 0.93 pi
-
Thanks Rainman!
I was getting frustrated playing with that grid.
-
K-Man is right
The sequence converges to $0, but even after including the effects of 93
heads in a row the Kasino is still slightly ahead -
Additional observations
We have established that B>F is required
G can be larger smaller or equal to A
B and F can be smaller or larger than G and A
For any set of values that fulfill all the problem requirements [say A=2, B=5, F=3, G=1: from k-man], you can add a positive number to B and F and/ or you can add a positive number to G and A and the new numbers will still satisfy all the conditions. So any of the following would still work:
- A=2, B=5, F=3, G=1
- A=12, B=5, F=3, G=11
- A=2, B=95, F=93, G=1
- A=12, B=95, F=93, G=11
The key to this is that:
- B and F must be relatively close in value
- Similarly G and A must be relatively close in value
Taking it further, the key to combinations that will work is the difference between B and F compared to the difference between G and A.
We have already established that B>F and that G can be larger, smaller or equal to A.
So to satisfy the problem
Abs [G-A] < [b-F].
If B-F = 1, then G = A must be true
if B-F = 2, then G +1 = A or G=A or G-1 = A must be true
...
-
You are right, I thought about it some more [even before I saw your post]
B must be > F
G can be less than, greater than or equal to A
-
Sorry for the delay. I was out of touch over the holidays.
A+B> F+ G??]
style="color: rgb(40, 40, 40);">How can G>A cause [4]
A+B>F+G?style="color: rgb(40, 40, 40);">
style="color: rgb(40, 40, 40);">
style="color: rgb(40, 40, 40);">The easiest way to demonstrate this is by
applying number values for each person.style="color: rgb(40, 40, 40);">Let’s use some undefined, normalized units [my
Engineering professors would be aghast].style="color: rgb(40, 40, 40);">
style="color: rgb(40, 40, 40);">Let’s say that C, D, E and H, I, J each have a
value of 10style="color: rgb(40, 40, 40);">
style="color: rgb(40, 40, 40);">For the four people in question (A, B, C, D)
let’s assign values (11, 10, 8, 12)style="color: rgb(40, 40, 40);">
style="color: rgb(40, 40, 40);">Rechecking the earlier
equationsstyle="color: rgb(40, 40, 40);">
style="color: rgb(40, 40, 40);">[3]
style="font-family: calibri;">A(11) +F(8) = 19 < B(10) +G (12) =
22style="font-family: calibri;">
style="color: rgb(40, 40, 40);">[4]
style="font-family: calibri;">A(11) +B(10) =21
> F(8) +G(12) =
20style="color: rgb(40, 40, 40);">And G > A
-
Although we have not assigned a monetary value to the package, it does have value!
The problem statement says that the preferred outcome is to
Live
Get $500
Get the package
Taking the package out of the caluclation will certainly simplify things, [and may return the same answr],
BUT I think it may lead to a faulty conclusion.The package doesn't affect the decision at all. Since getting paid $500 is preferred to getting the package, then going to work is always preferred to staying home as long as you survive. So, it only boils down to the probability of getting killed. This is, of course, assuming that if you get killed you don't get anything. Since the package is out of the equation, so is the $500. The question is still the same and it seems it boils down to your risk tolerance for getting killed. I don't see a mathematical answer to this question.
-
+g+c+d+e>+g+h+i+j,>
+g+c+d+e>+g+h+i+j,>Given
A+B+C+D+E > F+G++H+I+J [1]
and
C+D+E = H+I+J [2]
H+I+J = C+D+E [2a]
and
A+F < B+G [3]
Therefore
[1] – [2] yields
A+B> F+G [4]
[4] + [2a] yields
A+B+H+I+J > F+G+C+D+E [5]
[3] +[2] and [3]+[2a] yield
A+F+C+D+E < B+G+H+I+J, and [6]
A+F+H+I+J < B+G+C+D+E [7]
Comparing [3] and [4] tells us that
B>F
and/ or
G>A
Since either condition could cause this, we cannot reduce this further.Since either condition could cause this, we cannot reduce this further.
-
Question: Can we attach a monetary value to getting the package today??
I am not sure whether it affects the outcome or not, but it could.
-
From the problem statement, we get that Alice takes 8:01 for any complete
mile in the course [0.00 to 1.00, 0.10 to 1,10, ..., 25.2 to 26.2]. Therefore at
the end of- 1 mile her time should be 8.017 min [8:01].
- 26 miles her time should be 208.43 min [208:26]
- 26.2 miles her time should be 210.04 min [210:02]
While Bob's time should be 209.60 [209:36]
The only way for her to win is if she can speed up the last fraction of
a mile [where there are no more race markers] such that she can make up the
difference -- is this possible?? [this is a math problem, not a
realistic athletic question]. - 1 mile her time should be 8.017 min [8:01].
Splitting the pizza costs fairly
in New Logic/Math Puzzles
Posted