BMAD

yes. if the new firm joins the market, there are less customers for the incumbent

The manager of a firm is considering the possibility of entering a new market, where there is only one other firm operating. The manager’s decision will be based on the profitability of the market, which in turn heavily depends on how the incumbent firm will react to the entry. The incumbent firm could be accommodating and let the entrant grab his share of the market or she could respond aggressively, meeting the entrant with a cut-throat price war. Another factor that affects the revenue stream is the investment level of the entering firm. The manager of the firm may invest to the latest technology and lower his operating costs (low cost case) or he may go ahead with the existing technology and have higher operating costs (high cost case). The manager estimates that if his firm enters the market and the incumbent reacts aggressively, the total losses will be $7 million in low cost case and $10 million in high cost case. If the incumbent accommodates, however, the firm will enjoy a profit of $6 million in low cost case and $4 million in high cost case. Given that the incumbent will act in their own best interest, what should the manager of the firm do?

If you equate a word to the number of its letters, what do you have? One is three, which is five, which is four of course. So one is four. Two is also three, which we now know to be equal to four (see above). We know three is four and four is four. We’ve been through that. Five is also equal to four. Six is three, which is five, which is four. Seven is five, which is four… . . . Twentyfive is ten, which is three. And three is five, which is four. . . . Ninehundred is eleven, which is six. And six is three. Three is five, which is four. Even zero is four. How about words that are not numbers? Nothing is seven, which is five, and that leads to four. Universe is eight, which is five, which is four. You are three, which we now know is four. I am one, which also boils down to four. So you and I are four… And if someone tells you that you and I are one, well, that’s true also, because one is three is five is four! Well, apparently all is four. And that is true of course because all is three, which is four. Is there a counterexample?

never mind, i explained the wrong part. If it rained today then there is a 50% chance of rain again and 50% it won't

even chance of getting snow, rain, or pretty day

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Consider the following process. We have two coins, one of which is fair, and the other of which has heads on both sides. We give these two coins to our friend, who chooses one of them at random (each with probability 1/2). During the rest of the process, she uses only the coin that she chose. She now proceeds to toss the coin many times, reporting the results. We consider this process to consist solely of what she reports to us. Given that she reports a head on the nth toss, what is the probability that a head is thrown on the (n + 1)st toss? Now assume that the process is in state "heads" on both the (n - 1)st and the nth toss. Find the probability that a head comes up on the (n + 1)st toss.

The Land of Oz is blessed by many things, but not by good weather. They never have two nice days in a row. If they have a nice day, they are just as likely to have snow as rain the next day. If they have snow or rain, they have an even chance of having the same the next day. If there is change from snow or rain, only half of the time is this a change to a nice day. How is the weather 15 days from now if there is a equal probability of each type of weather today (probability of each type)?

So are we saying that x^2 + C condition is true for all conundrums but not all numbers where this is true for are conundrums? E.g. All squares have four equal sides but not all four equal sided shapes are squares.

There is a proof that i have that show's that it should be possible but what you present is something that i reference in my proof as well.

Care to share your solution.

This is a good start.

I marked it solved originally but just realized you didn't show equality yet. Sorry.

Show that, unless the towns are all collinear, it is always possible to choose three of them so that they form a triangle with maximum angle of at least 120°.

It wouldn't let me paste my random tables so here you go...

The first 2n positive integers are arbitrarily divided into two groups of n numbers each. The numbers in the first group are sorted in ascending order: a1 < a2 < ... < an; the numbers in the second group are sorted in descending order: b1 > b2 > ... > bn. Find, with proof, the value of the sum |a1 − b1| + |a2 − b2| + ... + |an − bn|. The table below is generated at random, in two stages. Firstly, a value of n in the range 2..20 is chosen at random. Then, n numbers in the range 1..2n are chosen at random. These numbers are sorted to form the sequence ai; the remaining numbers are sorted to form the sequence bi. For each i, the value |ai − bi| is calculated, and the sum is given. Here are some generated random table . Use the tables to formulate a conjecture about each pair, (ai, bi.) Prove your conjecture, and hence find the value of the sum.

A rhombus, ABCD, has sides of length 10. A circle with center A passes through C (the opposite vertex.) Likewise, a circle with center B passes through D. If the two circles are tangent to each other, what is the area of the rhombus?

Bonanova, Do you mean 2^2 = 3^2 + 5 ?

Three pumpkins are sitting on my front porch. They are spherical, of radii 4 ft, 3 ft, and 2 ft, and they are mutually tangent. I put a fourth pumpkin of radius 1 ft on top, in the hollow . . .How high off the porch is the top of the last pumpkin?

Find pairs of positive integers where 2a = 3b + 5. Prove that you have found them all.