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Thunder365
Riemann Hypothesis
Riemann Hypothesis


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#4021
Solution: x+y+z=9, which has 11C2=55 solutions over 6^3=216 ways. 55/216?

NP:Points A and B lie on a circle of radius 2 and center at O and the measure of angle AOB is 90 degrees. What is the area of the smaller region bounded by the circle and the chord joining A to B?
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PostPosted: Tue Oct 13, 2009 3:19 pm  Back to top 
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goldenboy1.618
Poincare Conjecture
Poincare Conjecture

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#4022
Solution
The triangle becomes an isosceles right triangle with legs on the radius of 2, and hypotenuse of 2\sqrt2. We see that the legs of the triangle create a region that has a quarter of the circle's area, or \frac14\times4\pi = \pi.

The triangle has area \frac12\times2\times2 = 2, so the area of the smaller region bounded by chord AB and arc AB is \boxed{\pi - 2}.


NP
Delilah and Deborah play a game in which they roll a fair 6-faced die. Delilah wins 45 cents if the top face is composite. Otherwise, Deborah wins 60 cents. What is the expected value won on any giver roll?

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PostPosted: Wed Oct 14, 2009 4:44 am  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#4023
The expected value for who?
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PostPosted: Wed Oct 14, 2009 8:14 am  Back to top 
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$LaTeX$
Riemann Hypothesis
Riemann Hypothesis


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#4024
Solution

Assume we want the money that Delilah wins.
So that is equal to \$.45\cdot2-\$.6/cdot4}=-1.5.
Thus the expected winning for Deborah is \$1.50.

Which is hopefully correct.... maybe


New Problem

Ligon Middle School has 1200 students. Each student takes 5 classes, and each teacher teaches 4 classes. Each class has 30 students, and 1 teacher. How many teachers are there at Ligon?

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PostPosted: Wed Oct 14, 2009 4:24 pm  Back to top 
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FlyAgaric
Hodge Conjecture
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#4025
solution (maybe?)
Here's how I did it.
Start with 1200 for the 1200 students. Each Student takes 5 classes and each class has 30 students so 1200*5/30=200. Each teacher teaches 4 classes so 200/4=50. There are 50 teachers...?


NP: One quiz is worth half of a test. Susan took two quizzes and scored 76 and 82. Then she took a test and scored 79. She has another test coming in a week and wants to raise her average to at least 82. What is the lowest score she could get on the next test?
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PostPosted: Tue Oct 20, 2009 1:15 pm  Back to top 
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Maybach
Navier-Stokes Equations
Navier-Stokes Equations


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#4026
Solution
We can take the average of 76 and 82, which is 79. 82*3 - 79 -79=
\boxed{88}

NP
If a coin is flipped 8 times find the probability there will be 5 heads and 3 tails.

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PostPosted: Tue Oct 20, 2009 1:58 pm  Back to top 
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Math Champion
Yang-Mills Theory
Yang-Mills Theory


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#4027
Solution

This is simply \binom{8}{5}\cdot\frac {1}{2^8} = \frac {1}{32}


NP

What is the unit's digit of the product of 1!\cdot2!\cdot3!\cdot4!\dots\cdot100!?

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PostPosted: Tue Oct 20, 2009 2:30 pm  Back to top 
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FlyAgaric
Hodge Conjecture
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#4028
Solution
Since there is at least one factor of 10 in the product, the units digit must be zero


NP
I am building a fence. The posts are 2 inches thick and the space between them is 5 in. If there is a post on each end and there are 25 posts, how long is the fence?

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PostPosted: Tue Oct 20, 2009 2:44 pm  Back to top 
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Maybach
Navier-Stokes Equations
Navier-Stokes Equations


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#4029
Solution
50+120=170

NP

Grrr........

Maybach got 26/30 on 2001 Chapter Sprint round due to careless mistakes. If the probability Maybach makes a mistake on each problem is the same, what is the probability Maybach waill get problems 27, 28, 29, and 30 wrong?

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PostPosted: Wed Oct 21, 2009 4:14 pm  Back to top 
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jin thynj
Poincare Conjecture
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#4030
\displaystyle \binom{30}{4}^{-1}

PostPosted: Thu Oct 22, 2009 6:24 pm  Back to top 
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jin thynj
Poincare Conjecture
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#4031
NP: I have some red sweets, and you have some blue sweets. Find max{|S|} if S is the set of all possible couples (red sweet, blue sweet) Rolling Eyes

PostPosted: Thu Oct 22, 2009 6:35 pm  Back to top 
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cgyao15
Navier-Stokes Equations
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#4032
hmm... how many sweetw do i have and how many does u have?
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PostPosted: Fri Oct 23, 2009 2:29 pm  Back to top 
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Thunder365
Riemann Hypothesis
Riemann Hypothesis


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#4033
Okay, NP:

Find the number of solutions to the equation 2a + 5b = 2009 for which both a and b are positive integers.
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PostPosted: Fri Oct 30, 2009 10:56 am  Back to top 
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MoTheMan
Poincare Conjecture
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#4034
We count 2s:
2 2s and 401 5s
7 2s and 399 5s
..
1002 2s and 1 5
That is 201 solutions.

NP:
Abe and Bob win a Banana Creme-Chocolate Mousse pie worth 18 dollars at the fair. Half of the pie is one flavor and the other half is the other flavor. Abe likes Banana 5 times as much as chocolate and Bob likes Chocolate twice as much as Banana. If Abe cuts it so that he likes each piece the same amount, and one piece is all Banana, and Bob picks the piece that is worth most to him, what value does Bob put on his piece?

PostPosted: Thu Nov 05, 2009 5:30 pm  Back to top 
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FlyAgaric
Hodge Conjecture
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#4035
My interpretation:

It is cut into thirds (one banana, one chocolate, one both). Bob picks chocolate because that is his favorite. Since he has 1/3 of the pie, it is worth 6 dollars. However the problem was not very clear.

If Bob thinks chocolate is twice as valuable as banana then he thinks that the chocolate half is worth 12 dollars to him (?) so his slice would be worth 8 dollars to him (it is 2/3 of the chocolate half). I'm not sure though. Huh?
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PostPosted: Fri Nov 06, 2009 4:26 pm  Back to top 
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MoTheMan
Poincare Conjecture
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#4036
I am sorry. Let me make the problem more clear.
Abe cuts in into two pieces. One piece is a certain amount of banana and no chocolate. The other is all the chocolate and the other bit of banana.
Abe thinks the banana half is 15 dollars and the chocolate half is 3.
Bob thinks the banana half is 6 dollars and the chocolate half is 12.

PostPosted: Sat Nov 07, 2009 5:48 am  Back to top 
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FlyAgaric
Hodge Conjecture
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#4037
Ok then

Abe wants both slices to be worth 9 dollars. The all-banana slice is 3/5 of the banana half or 3/10 of the pie. Bob takes the half that is 7/10 of the pie which is worth to him 18-3.60=14.40 dollars.
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PostPosted: Sat Nov 07, 2009 7:17 am  Back to top 
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steve123456
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#4038
Post a new NP after you solved one. I'll post this new problem.

NP: Max is on his triangular lawn in his backyard. He labels the vertices of the triangle A , B , and C, with AB=10, BC=15, and AC= 13. Then he draws the angle bisector of A until it intersects with BC at point X. What is the length of AX?

PostPosted: Sat Nov 07, 2009 10:34 am  Back to top 
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