BMAD

A dog hiding in the tall grass at the edge of a circular field spots a fox at the center of the eld. The fox, unable to see the dog but sensing the danger, begins to run at a constant speed along a straight line path towards the safety of a uniformly selected random point at the edge of the eld. At that same instant the dog jumps the fence and runs straight at the fox at a constant speed M times that of the fox. Find the probability that the fox will escape the field, and the value of M at which the dog must run to have a 50% chance of catching the fox.

Stocks are only worth what someone is willing to pay for it. So in this case, wealth will be determined by your dollar amount remaining and the price the market will pay for your remaining stocks To be clear though, market purchase price will be run 1 additonal time based on our last day of sales. So final wealth will not be determined based on THESE (day 10) numbers.

Stocks are only worth what someone is willing to pay for it. So in this case, wealth will be determined by your dollar amount remaining and the price the market will pay for your remaining stocks

Three men: Fermat, Galois, and Hilbert, decide to fight a pistol duel. They'll stand at the corners of an equilateral triangle, and each man, in order, will aim and shoot wherever he pleases. They choose randomly who will be shooting first, second, and third, and will continue in order until two of them are dead. All three know Fermat always hits his target, Galois is 80% accurate, and Hilbert hits his mark half the time. Assuming that all three adopt the best strategy and that nobody is killed by a wild shot not intended for him, who has the best chance to survive, and why? Find the survival probabilities for each man.

Herman knows how old he is turning this birthday; you don't. He is as many years old as the largest number of divisors of any integer N less than or equal to 20,000. How old is Herman turning, and what's the smallest such N?

Fearless Frank decided to play a fair coin-flip game with probability 1/2 of winning each bet, and risked 1/m of his fortune (originally A dollars, m>1) at every flip. After 2n games, Frank has won n games and lost n. Choose and explain the correct answer from this list: a) Frank has broken even; he still has his A dollars b) Frank is predictably ahead by a certain amount c) Frank is predictably behind by a certain amount (in case b or c give the exact formula in terms of m and n) d) Frank is now ahead, behind, or even, depending on the order in which the wins and losses occurred e) He is ahead, behind, or even, depending on m and n

i don't believe there is a next word

your risks definitely paid off. Your strategy was an excellent strategy for short term stock trading. In the long run, diversification is preferred as you are not whole dependent on the success of a single stock, nor do you lose everything if that stock becomes worthless. It isn't too late though. So don't be careless!

Day 10: This is the final day to make any last second changes. Current wealth totals are listed. Let the last second scurrying begin.

After searching another forum, the other site said you need to find a combination somewhere for the door then turn the dials to make the combination of numbers. The center number changes to match the combo you want. I would try: 53 42 35 33 41 Apparently a lot of people are having trouble finding the combo. No gurantee it will work, just what someone posted on a forum.

Yay!

I have no idea what I am doing but I will give it a try.

Grail

trawl

How many lines are needed to make exactly N squares? Solve for 1<=N<=15. For example, to make five squares, six lines are needed (draw a 2x2 grid, to give four 1x1 squares and one 2x2 square.) But to make exactly four squares, seven lines are needed.

Given a compass, a straightedge, and a line segment AB of length 3, how could you draw a line segment with a precise length of the square root of 3?

Day 9

nice. i was trying to make it say 53 :-)

on the right track

For a ripple shuffle, it would take more than two shuffles to gurantee a random situation (every possible orientation of cards) is possible Cheers!

I am referring to "riffle shuffles". The random riffle shuffle is modeled by cutting the deck binomially and dropping cards one-by-one from either half of the deck with probability proportional to the current sizes of the deck halves. How many shuffles must be done, to where every possible card configuration is possible?

Assume the 52 card deck is in order at the beginning. 1,2,3,4,5,6,7,8,9,10,... how many shuffles would it take to truly break the order up

Draw three triangles, such that each overlaps a different vertex of the other two. That is, in triangles A,B,C with vertices (A1,A2,A3), (B1,B2,B3), (C1,C2,C3), you'll find B1 and C2 inside triangle A, C1 and A2 inside triangle B, and A1 and B2 inside triangle C. Can this be done with isosceles right triangles? What is the smallest internal angle possible in the triangles? What is the smallest rectangle which holds all three triangles?

Team A, Team B, Team C, and Team D decided to play a small football tournament against each other; each team would play once against each other team. After all the games were played, Team A won all three of their games, scoring six goals total while only giving up one goal. Team B won a single game, tied a game, and lost a game. They only scored two goals total, while giving up four goals. Team C won a single game and lost two games. They only scored two goals total, while giving up two goals. Team D did not win any games; they tied once and lost twice. They only scored two goals, and gave up five goals. If you take the number of goals Team A scored against the Team B, and you multiply that by the number of goals that Team C scored against Team D, and you multiply all that by the total number of goals scored in the game between the Team A and Team D, and you add to that the number of goals Team C scored against the Team B, what number will you get?