> In a financial arrangement you are promised $1000 the first day and > each day after that you will recieve 35% of the previous days amount. > When one days amount drops below $1, you stop getting paid from that > day on. What day is the day you would receive no payment and what is > your total income? Use a formula for the nth partial sum of a > geometric sequence. The last sentence pretty much sums up the advice we'd give you. Figure out on what day the amount drops below $1. Look up (or derive, if you are advanced enough) the formula for the nth partial sum of a geometric sequence. Put the two parts together to get teh final answer. -- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Professor of Computer Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics Affiliations for identification only. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > if i reviewed 30 charts and 13 were compliant what would be the > percent. Please show me how to calculate. Cindi: The answer is 100 * 13/30 or 130/3. You can do the division yourself. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== If I'm looking to list all possible 4-digit combinations using 1 thru 6 (i.e. 6^4 = 1296 possible combinations), is there a way by which MS Excell can be made to produce such a list, please? -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== Author: Graham 6 (i.e. 6^4 = 1296 possible combinations), is there a way by which MS >Excell can be made to produce such a list, please? Graham, First, I question the wisdom of such an assignment; your task seems to me too much like busy work, but OK, let us accept it. Second, there may well be some Excell function that does the job or some other really clever way to bend Excell to your will, but I do not know what it is. But, until you get a better answer, here is my suggestion. Write out the numbers, not as complete numbers, one 4-digit number in a cell, but put each place value in a separate column. Here is what I mean. First, make the column widths as narrow as possible. You can do this by modifying the sheet's properties so that the default column width is 1 or 2 (1 may not show the numbers). Second, you know that the 4th digit of your numbers will repeat in blocks of one to six, down the page, so write out the 4th column 4th col ------ 1 2 3 4 5 6 1 2 3 4 5 6 etc. You can copy and paste this column as many times as you need. Next, for each block in the 4th column, the third column will repeate the same digit and, as you go down to the next 4th-digit block, you increment the digit. Here is what the 3rd and 4th blocks look like: 3rd 4th 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 and so on. As you work with this, you will speed your workd by discovering clever ways to cut and paste. I hope someone will give you a better answer. In the meantime, I think this approach will work faster and easier than actually trying to type out the permutations. Good luck, Haim -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== << If I'm looking to list all possible 4-digit combinations using 1 thru 6 (i.e. 6^4 = 1296 possible combinations), is there a way by which MS Excell can be made to produce such a list, please? >> What could be it good for? Is it not much easier to generate it directly? For two numbers: 11 12, 21, 22 For three: 111 112 113 121 122 123 131 132 133 211 212 213 221 222 223 231 232 233 311 312 313 321 322 323 331 332 333 -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > > The natural numbers don't constitute a field either. Nor do they form > a ring. But their algebraic structure is very similar to that for the > power set of a given set X, with the operations of union and > intersection. The operations are closed, associative, and commutative. > The main difference is that there are two distributive laws instead of > one. But, I have seen in several books that intersection precedes > union in the order of operations. It's arbitrary. If we were inclined > to write all polynomials (over an appropriate ring) in factored form, > we would perhaps choose to perform additions first. > > Let's stick to the natural numbers if you wish, and so examine the > argument given by Ron. (I wanted to start with the field structure, he > with the natural numbers. Maybe he was right to do so. According to > Kronecker, the natural numbers are the work of God, but everything > else is the work of Humanity.) > > I do not see how a decision is arbitrary when one choice repeatedly > leads to a contradiction and the other never does. > > Let multiplication in the natural numbers be understood as repeated > addition of the same number: > > a + a + . . . + a (for n instances of a) > = (a*1) + (a*1) + . . . + (a*1) (for n instances of a*1) > = a*(1 + 1 + . . . + 1) (for n instances of 1) > = a*n > > Then giving default precedence to addition repeatedly yields > contradiction, but giving default precedence to multiplication never > does. For example: > > 3 + 5 + 5 = 13 > > 3 + 5 * 2 = 16 (addition gets default precedence) > > 3 + 5 * 2 = 13 (multiplication gets default precedence) Where's the contradiction? If addition were to have default precedence, we would have 3 + 5 * 2 = 16. So, 3 + 5 + 5 would have to be written as 3 + (5*2). We could no longer substitute (within a sum of different numbers) a sum of n equal addends m with the expression n*m. We would substitute (n*m). I don't see the contradiction. I will admit it is CONVENIENT to omit the parentheses, and give preference to multiplication. But I don't believe it is absolutely required for any mathematical reason. > To anticipate: > > Of course we can give precedence to addition if we want, such as in (3 > + 5) * 2 = 16, but this is about default precedence. On expressions > involving both addition and multiplication with no parentheses > denoting precedence, this is about having a default precedence grammar > that avoids contradiction. > > Cordially, > > Paul -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > My son is currently in Algebra 2 and he can't figure out the following > problem (nor can I): > > sqrt(14)/(sqrt(6) + sqrt(7) - 1) > The goal here is to eliminate radicals in the denominator. Multiply both the numerator and denominator by sqrt(7) - [sqrt(6) - 1] This will leave 2*sqrt(3) in the denominator. Multiply both numerator and denominator by sqrt(3) Then simplify. I get [7*sqrt(3) - 3*sqrt(14) + sqrt(21)]/6 for the final answer. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > sqrt(14)/(sqrt(6) + sqrt(7) - 1) Is there a typographical error here? Maxima cannot find a simpler expression for this. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html