When using this rule, why must k be greater than one? Provide an example of its usage. Can anyone help me with this question? :-(
201889
Do you mean Chebychev's theorem of proof of Bertrand's postulate on prime number distributions, or Chebychev's theorem concerning standard deviations? \n\nEdit your question to clarify and I'll answer more...\n\nOk, Chebychev's theorem states that at least 1-(1/k^2) of a distribution lies within k standard deviations of the mean.  The reason k must be more than 1 is that at that low of a standard deviation exactly what type of distribution you have becomes important.  Chebychev's theorem works for any distribution, so it can be used to assess general population size, at least within a fairly wide range, with less analysis using just the sample standard deviation.  So, it works for sparse data, non-normal distributions, and so on, within its limits of only holding for calculations of what lies within a range larger than one standard deviation.  \n\nSo, for example, you can take a small subset from a set of data with a mean of 50, and a *sample* standard deviation  of 8, and from Chebychev's theorem know that if you were to take a larger sample, at least 75% (1-1/2^2) would lie within the range of 44 to 66, that is, 50+/-16.
450032
Ok, your question makes absolutely no sense.  Why say we can only use 1 through 16, and then say we can only use 1?\nSome ways, assuming I translated your question correctly, would include 16 + 16 +1 +1, and 16 + 15 + 3.... There are almost an infinite number of ways to add to 34 with the numbers 1-16, so please rephrase your question.
15662
There are MANY faults in this reasoning.\n\nA) A finite number can only be divisible by integers that are SMALLER than the given number. In the first example, 720 is NOT divisible by 721, 722, etc. In the second example, 1,000,000,000,000,000,000! is not divisible by 1,000,000,000,000,000,000! + 1.\n\nB) A much easier way to find out the factors of a number is to factor it: for example: \n720 = 72 x 100 = 9 x 8 x 10 = 3x3 x 2x2x2 x 2x5, then you can use a method in combinatorics to find out how many different numbers you can make by multiplying the factors of 720 in different ways to see how many different numbers go into 720.\n\nC) Combining arguments A and B, you CAN show that 1,000,000,000,000,000,000! IS divisible by all integers smaller than 1,000,000,000,000,000,000. (I do intend to leave out "!")\n\nD) Finally, and most importantly, if you are going to show something is true for all integers you must either sit down for all time testing the validity of your hypotheses on EVERY NUMBER or you can use Mathematical induction.\n\nInduction means that you show your hypothesis is true for some initial number n (e.g. show that 720 is divisible by 1). Then you must show that IF your hypothesis is true for any n, THEN it must be true for the next number, n+1. You must carry out this argument WITHOUT REPLACING n WITH A NUMBER!\nYou should try to prove that1,000,000,000,000,000,000! is divisible by all positive integers smaller than 1,000,000,000,000,000,000 by showing if it's true for n, then it's true for n-1.
242835
m > 42.\n\nm - 13 is greater than 29. \n\nSo you add the 29 and the 13.\n\nso m must be greater than 42, hence the answer m > 42.
297938
Above answer is ok except for not using the -- operator:\nNeed to start at 10 and to decrement i use the statement:\n\ni--;\n\nThis changes the problem somewhat.  Another problem in the first answer is that fact is not initialized, so it will always equal zero resulting in a division by zero error.
293391
=IF(E12=0,0,\nIF(AND(E12>0,E12<4),\n5.9,IF(E12>=4,12.9,\n"negative value")))\n\nAll on one line!\n\nEdit:  The >= (greater than or equal to) is required else the value of exactly 4 would return "negative value"
425593
You are very close. If K = -0.032 is accurate enough, then use it.\n\npA = Ae^(19 * -0.032)\np = e^(19 * -0.032) ~ 0.544\n\nIf you leave K = ln(.5)/22 then\np ~ 0.5496
329601
because a+b-c = 0, \nu cant cancel out (a+b-c) from both sides \nu can only cancel if (a+b-c) is not zero, which is not the case. \n\nhint: i should get best answer
144883
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
105621
I'll assume you mean y=a e^(kx) +b e^(-kx). Then dy/dx=ak e^(kx)-bk e^(-kx). Set this to 0 and do some work to see that\ne^(2kx)=b/a, so x=[1/(2k)]ln(b/a). If b/a <0, there is no critical point. If a<0 and b<0, the critical point is a maximum, not a minumum. So, for a>0, b>0, we get the minimum occurs with the x value above. Plug in and use properties of exponents and logs to find that the minimum value is f(x)=2sqrt(ab).
158685
Use the ratio test. We want the absolute value of the ratio as k approaches infinity of the (k+1)th term to the kth term to be less than 1. So take the expression x^n/n^(1/5) and let n = k + 1. Stick the result in the numerator. Then let n = k and stick that result in the denominator. Now take the limit as k goes to infinity. You should just end up with x (I don't want to write the whole process out here cuz it's hard to write equations but you should get something that looks like x[k/(k+1)]^(1/5) which approaches x as k -> infinity). So if |x| < 1, you're good to go. If x is plus or minus 1, we're still not sure if the series converges or not. The x = 1 case gives us a p-series with p = 1/5, which diverges. The x = -1 case gives us an alternating series in which the magnitude of each term is always decreasing and approaches zero as n goes to infinity. Thus, it passes the alternating series test for convergence and the series converges for x = -1.\n\nSo the interval of convergence is [-1,1) and the radius of convergence is 1.
278593
In line 5, you are dividing by x-y. Since x = y, then x - y equals zero.\n\nYou can not divide by zero.
558325
To define an equation, you can use the syntax\nf := x -> (k*x*(1-x));\nWhich is creating a function defined on a variable x.\nThen, you can set the value of k by saying \nk := 2;\nIf you then say\nf(.1);\nit will give you the value of f(.1).  You can repeat this with the value you get from this.
463723
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
367961
x+78 [greater than or equal to] 120\nx [greater than or equal to] 120-78\nx [greater than or equal to] 42
639373
When the book asks for a rule, they usually want it in a formula so it can be used no matter what the value of x is. \n\n1) Since the value of 'y' has a steady decrease of 2, then we know if x=0, y=87.  \n2) Start your formula: "y = 87...". \n3) We know that we'll have to be subtracting something since the value is decreasing: "y = 87 - ..."  \n4) Plug in what you already know, using something other than x=1, to figure out what more needs to be done (I'll use x=3):  87-81=6. \n5) How does the value of 'x' relate to this new figure? 3*2=6. \n6) Check to see if multiplying the value of 'x' by 2 will make up the difference after 87 is subtracted from the value of 'y'. \n\nThus we know that in order to find the value of 'y' we must first start with the number 87 and then subtract whatever we get after multiplying x*2 (i.e. 2x). \n\nAnswer: y = 87 - 2x
536025
ry +s = tx - m\nYou must leave y alone for indicate its value in terms of other variables.\n\nFirst of all you move the s to the other side or the = as shown: ry =tx- m - s (you change the sign of s)\n\nThen you move the r as follows: y=(tx-m-s)/r\n\nAnd thats it \n\nYou have a restriction because the r has to be different than 0 to avoid a division by zero.  So restriction is r different from zero.