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Thunder365
Yang-Mills Theory
Offline Joined: 27 Feb 2009 Posts: 500 Location: Michigan
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Solution: , 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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Posted: Tue Oct 13, 2009 3:19 pm
goldenboy1.618
Poincare Conjecture
Offline Joined: 14 Mar 2009 Posts: 145 Location: USA
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Solution The triangle becomes an isosceles right triangle with legs on the radius of 2, and hypotenuse of
. We see that the legs of the triangle create a region that has a quarter of the circle's area, or
.
The triangle has area
, so the area of the smaller region bounded by chord AB and arc AB is
.
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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Posted: Wed Oct 14, 2009 4:44 am
AwesomeToad
Yang-Mills Theory
Offline Joined: 25 Apr 2009 Posts: 787 Location: West Lafayette, IN (Rating: 1337)
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The expected value for who?
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Posted: Wed Oct 14, 2009 8:14 am
$LaTeX$
Riemann Hypothesis
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Solution
Assume we want the money that Delilah wins.
So that is equal to
.
Thus the expected winning for Deborah is
.
Which is hopefully correct....
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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Posted: Wed Oct 14, 2009 4:24 pm
FlyAgaric
Hodge Conjecture
Offline Joined: 31 May 2009 Posts: 85 Location: Oort Cloud
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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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n. 1. type of mushroom which is bright red with small white dots; consumption may lead to death or extreme, short-term growth
Posted: Tue Oct 20, 2009 1:15 pm
Maybach
Navier-Stokes Equations
Offline Joined: 06 Jun 2009 Posts: 1651 Location: NW Indiana
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Solution We can take the average of 76 and 82, which is 79. 82*3 - 79 -79=
NP If a coin is flipped 8 times find the probability there will be 5 heads and 3 tails.
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Posted: Tue Oct 20, 2009 1:58 pm
Math Champion
Yang-Mills Theory
Offline Joined: 26 Jun 2009 Posts: 562 Location: Redmond, WA
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Solution
This is simply
NP
What is the unit's digit of the product of
?
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Posted: Tue Oct 20, 2009 2:30 pm
FlyAgaric
Hodge Conjecture
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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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Posted: Tue Oct 20, 2009 2:44 pm
Maybach
Navier-Stokes Equations
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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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Posted: Wed Oct 21, 2009 4:14 pm
jin thynj
Poincare Conjecture
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Posted: Thu Oct 22, 2009 6:24 pm
jin thynj
Poincare Conjecture
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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)
Posted: Thu Oct 22, 2009 6:35 pm
cgyao15
Navier-Stokes Equations
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hmm... how many sweetw do i have and how many does u have?
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Posted: Fri Oct 23, 2009 2:29 pm
Thunder365
Yang-Mills Theory
Offline Joined: 27 Feb 2009 Posts: 500 Location: Michigan
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Okay, NP:
Find the number of solutions to the equation for which both and are positive integers.
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Posted: Fri Oct 30, 2009 10:56 am
MoTheMan
Poincare Conjecture
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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?
Posted: Thu Nov 05, 2009 5:30 pm
FlyAgaric
Hodge Conjecture
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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.
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Posted: Fri Nov 06, 2009 4:26 pm
MoTheMan
Poincare Conjecture
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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.
Posted: Sat Nov 07, 2009 5:48 am
FlyAgaric
Hodge Conjecture
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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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Posted: Sat Nov 07, 2009 7:17 am
steve123456
Hodge Conjecture
Offline Joined: 01 Nov 2009 Posts: 77 Location: Why would you care? Hmm, fine I live in Pleasanton,CA
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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?
Posted: Sat Nov 07, 2009 10:34 am
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