What Is The Probability That Exactly N-2 Letters Are In The Correct Envelopes?

From Home Schooling Forum:

What is the probability that exactly n-2 letters are in the correct envelopes? If there are n different letters to be sent out to n different people and you're in a rush so you randomly put the letters into envelopes. What is the probability exactly n-2 letters are in the right envelope?

Answer: This is a tricky question, because the trials are not independent. However, the calc"ulation is simple. I assume you know what n! means, "n factorial," n * (n-1) * (n-2) *.......3 * 2 * 1, with 0! defined as 1. So 2! is 1*2, and 3! is 3*2*1? and so on. Our random variable X will represent the number of letters in the correct envelopes. We want to find P(X=n-2). I will write this just as P(X) unless I use a specific value. We will do this for n = 1, 2, and 3 by hand, then use a formula for n > 3. ' If n = 1, P(X=-1) is equal to 0, obviously. With only one envelope and one letter, you can't mismatch them. If n = 2, then n-2 = 0, and there is a .5 chance of getting 0 letters in the correct envelope. Given envelope 1, you have an equal chance of putting letter 1 or letter 2 in it. You will then out the other letter in the other envelope. So you will either have 0 or 2 letters in the wrong envelope, with a 50-50 chance of each. If n = 3, then n-2 = 1. One letter will be correct, and the remaining pair will be mismatched. So the correct one can be either 1,2,or 3: the remaining pair will be mismatched. So there are three ways you can have X = 1, and there are 3! ways of arranging 3 letters in 3 envelopes. So P(X) again equals .5. Notice in all these cases, we can have n-2 correct only if we have exactly 2 incorrect. In general, for numbers greater than 3, we calculate the probability like this: We have n! ways of assigning n letters to n envelopes. (because: We have n possibilities for the first letter, n-1 for the next, n-2 after that is chosen . . . . 2 ways to choose the second to last, and only one last letter, which goes in the last envelope. If we multiply these ways out, we get n!) If we can figure out how many ways we have of choosing exactly one incorrect pair, we're done. That's easy. Suppose we choose the envelope first. We have n possible ways of choosing it. Then we choose the letter. We have n-1 ways of choosing the letter so that we have an unmatched pair: we can choose any letter, except the one that properly belongs to the envelope we have chosen. So, since the probability of choosing any envelope or letter is the same, we can get the probability from the relative frequency: For n > 3, P(X=n-2) = n*(n-1)/n! Dividing the numerator and denominator by the numerator reduces this to 1/(n-2)! Remember that the results for n < 4 are separate. If n = 1,

What Is The Summation Of N Raised To N For Finite N?

From Mathematics Forum:

What is the summation of n raised to n for finite n? I am interested in finding the summation of n^n for finite n. Tried a bit of different methods for finding its value, but can't seem to get a hand on it. I also tried searching on go ogle and all kinds of different links, still no clue. If anyone knows anything about this, please help me. A relevant web resource would also do. Thanks in advance.

Answer: I do not know any general polynomial or other closed function equivalent and suspect that there is not one. Of course you know Sigma (n^n) => 1 + 4 + 27 + 256 + .... Perhaps you only need the first ten finite sums, which are:- 1, 5, 32, 288, 3413, 50069, 873,612, 17,650,828, 405,071,317 and 10,405,071,317 The last result is 3.40314156939*Integral [x^x] from 0 to 10, but so what ? There is a tedious series expansion for Integral [x^x] about x = 0 which Wolfram can supply. It starts as I = x + (1/2)x^2ln x, and continues with ridiculous complexity through x^8 terms and ... That is no use for calculating say the Th finite sum Any particular definite integral of x^x can, of course be calculated to an arbitrary degree of accuracy by numerical techniques. Any finite sum can also be found. Sorry, but I think what you want does not exist. Regards - Ian

How Long Will It Be Before The N Word Is Made Legal For Public And Private Sector Places Of Work?

From Current Events Forum:

How long will it be before the N word is made legal for public and private sector places of work? Blacks, Latinos say the N word with ease. Whites think it, even if they are married or have sex with blacks. Blacks have made use of the N word easy. Arabs, whom blacks admire, say it with ease.

Answer: But there are two "n" words. One starting Ni the other NE The first one is very offensive.Has been for many years The second one is not now used. I cant see any of them coming back for use by white people ?

How Do I Connect To The Bel kin Pre-N Router Using The N Frequency Instead Of G?

From Computer Networking Forum:

How do I connect to the Belgian pre-n router using the n frequency instead of g? I have a dell cps that comes with a wireless draft-n card. I just can't figure out how to connect that to my Belgian pre-n using the n frequency so I can get greater range and speed. My setting on the Belgian is set to either mixed, b-only, or g-only. I set it on both or g-only and it's still only the g connection. Please help.

Answer: You can not connect at N speeds the wireless card that you have will connect to B, G, or N routers but the Bel kin router only knows about B and G . Your only solution is to purchase a 802.11n router

What'S The Difference Between An N Router And An N Router With 1 GB With 4 Ports?

From Computer Networking Forum:

what's the difference between an N router and an N router with 1 GB with 4 ports? Which one do I need? The regular N router is less expensive than the later, but I want to buy the right one. What is benefit to the router with 1 gb?

Answer: The GB is the speed of the Ethernet ports on the router. If your machines all have mobs network cards then GB will allow you to have mobs Ethernet connections. The real truth these days about N etc is what do YOU need the system to do. If all you want is Internet access.. even a G router will work well for you as mobs (G speed) is still way faster than your Internet connection! So take a look at your local in house use. Do you have multiple computers or just one or two? Do you constantly share files between those? Do you expect to need a server? Is your main use for Internet, if so I would get a good G router with good antennas and save my money for something useful. If you need local machine to machine speed then N and mobs network cards etc is a requirement! Its important to note - neither of these will make your INTERNET faster just your local network! Your USP controls your Internet speed.

Let N Be An Arbitrary Integer. What Is The Least Number Of Weights That Can Be Used On?

From Mathematics Forum:

Let N be an arbitrary integer. What is the least number of weights that can be used on? Let N be an arbitrary integer. What is the least number of weights that can be used on a scale pan to weigh any integral number of pounds from 1 to N inclusive, if the weights can be placed in either of the scale pans?

Answer: What you need to figure out is how much "freedom" you have. First think of a simple application: If you only have to count in positive ways, how many stones do you need? Ignoring the other side of the scale, what would the solution be? Well, I want to be able to form all numbers from 1 to N, using some "system" -- what system would you use? How about a number system? For example, base 10 or base 7 or even base 2. The point is that if you have a number of pounds k (1 ≤ k ≤ N), you can write k in many ways, including sums of particular weights you might (or might not) have. If I had the weights {1,2,4,5} and I wanted to weight this against 7 pounds, I could either use:       7 = 5 + 2       7 = 4 + 2 + 1 So you see -- we don't want to be "redundant" if we are going to minimize the weights we use. To use as few as possible, we can use binary. The weights {1,2,4,8,16} for example can weigh anything from 1 to 31 with only the four of them. But you ARE allowed to use the other side of the scale. For example, to weigh out 6 using the same example as above, you could do:       6 = 4 + 2       6 = 5 + 2 - 1       6 = 5 + 4 - 2 - 1 So you see, there are many ways to do this. What you want to do is "build up" the system and see if you can guess the pattern. In the old system, you started with 1 and you couldn't reach 2, so you added 2. Then you couldn't reach 4, so you added 4, etc. However, if you are allowed subtraction, then here's how to do it. Start with a weight 1 stone. I cannot reach 2. Should I add 2? NO. Whatever stone I add, if it is "k" then I can reach k-1. So the k-1 should be 2. In other words, k=3. Now I have 1 and 3. I can reach 4, but not 5. Do I add 5? No, I add 5 + [total] = 5+4 = 9. Now I can reach 1 + 3 + 9 = 13 but not 14. I should not add 14 to my set. I should add 14 + 13 = 27. From there, it continues. 81, 243, etc. The answer is: powers of 3. You need only roughly log[base 3] ( N ) to do this.

What Is The Maximum Number Of Electrons That Can Be In The N Shell Of A Period 4 Element?

From Chemistry Forum:

What is the maximum number of electrons that can be in the N shell of a period 4 element? I know the N shell can hold a max of 32 electrons, but what about when it is just a period 4 element? I think the outermost shell of period 4 elements is the N shell, so there must be some special rule.

Answer: Each level follows the rule e=2n^2 so level 1(n=1) e=2 level 2(n=2) e=2*4=8 n=3 so e=2(9)=18 n=4 so e = 2(16) n=5 so e=2(25)=50 ............

How Do The Fermi Levels Of P And N-Regions Of An Unbiased P-N Junction Adjust To Become Equal?

From Physics Forum:

How do the Fermi levels of p and n-regions of an unbiased p-n junction adjust to become equal? The original Fermi level of the p and n-regions taken separately are different due to difference in concentration and type of doping. However, at the junction, the energy level diagram usually depicts a common Fermi level while conduction and valency bands are displaced. How does that happen?

Answer: Energy levels in materials are always relative. In other words, you can say that the conduction band is so many Ev above the valence band, but it has no particularly absolute value. The Fermi level is simply the energy level of the topmost occupied electronic state (at T = 0). Th id once again is relative - you can say where it is with respect to say the valence band. Until, that is, you join the two materials. When you do this the energy of the topmost occupied state on the n side *must* be the same as that on the p side because the materials are in contact. If it were not, then electrons would flow into the lower energy states across the boundary until they were. Clearly, for these two levels to be the same across the boundary, the energy bands measured relative to the Fermi level must change, because the band structure must be continuous across the boundary. For this to happen the conduction and valence bands must curve. The origin of this curvature is, of course, and electrical potential energy gradient across the junction. This is what gives rise to the intrinsic junction bias.

How Can I Show That The Function F(N)=2N^3+N^2+6N+3 Always Produces A # Divisible By An Odd # Greater Than 1?

From Mathematics Forum:

How can I show that the function f(n)=2n^3+n^2+6n+3 always produces a # divisible by an odd # greater than 1? I tried to solve it by substituting the n value as 3 and solve for the y-value. Can someone tell me how can I solve this problem. Please!

Answer: One way would be to try to factor the polynomial formula for f. And this is successful: f(n) = (2n + 1)(n^2 + 3) for all n. (Check by expanding it out) So in fact f(n) is always divisible by 2n+1 (which is odd and greater than 1 for all n >= 1.) Note that you don't need to know where the factorization "comes from" to be able to understand how it proves the statement. If you were curious, however, I found the factorization by looking for roots of the polynomial f(x) = 2x^3 + x^2 + X + 3. I know that if I find one, r, the polynomial factors as (x-r) * (a quadratic). The "rational root theorem" (see link below) suggests trying numbers p/q with p = 1, 3, -1, or -3 and q = 1, 2, -1, or -2. When I did this I found that -1/2 was a root of f(x). This tells me that f(x) factors as (x + 1/2) * something else, or if I put in a factor of 2, that f(x) factors as (2x + 1) * (2 something else). Replacing x with n above, I can do polynomial division by 2n+1 and see that the division comes out with no remainder.

How Long Will I Get To Receive My Receipt After I Filed N-400 Application For Naturalization?

From Immigration Forum:

How long will I get to receive my receipt after I filed N-400 application for Naturalization? I applied for N-400, Application for Naturalization Test. I have not received a receipt yet but I am just wondering when will I receive it and how long it takes for the application to process. I live in California.

Answer: It took me about a month to receive a letter of receipt that the INS received my N-400 application. Then it took about another 6 weeks before I received a letter to report for biometrics. Then about 3 months for a letter to report in Federal court for ceremony. Now, please remember, that this is how long I waited, it does not mean that you will have to wait this time--the time in which you wait may be shorter or longer. Good luck to you!

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