analyze two simple problems? 1. If m men paint a house in d dayz how many dayz will m+2 men take?
461412
1) assuming it's inversely proportional then you have multiplied by m+2/m so it's dm/(m+2)\n2)(m+2)d/m
718274
4 walls -> 5m x 2.5m x 4 = 50 sq.m\nRoof    -> 5m x 5m       = 25 sq.m  total 75 sq.m\n\nLess for 1 door -> 0.9m x 1.8m = 1.62 sq.m\nless for 2 windows 1.2m x 1m x 2 = 2.4 sq.m total 4.02 sq.m\n\nArea to be painted 75-4.02 = 70.98 sq.m\nNow just multiply with the rate
322820
mmh... thats a tough question.
542346
haha good question :]\nwe're just studied those in math, and a lot of our problems had "real-life" situations, so i'll just give you some examples of them.\n"A farmer has at most 80 acres for which to plant two crops, corn and wheat. Planting and harvesting cots, for which no more than a total of $12,000 is available, are $300/acre for corn and $100/acre for wheat. How many acres of each crop would you suggest he plant?"\n"Ken is planning a picnic for the youth orchestra. He is going to make chicken sandwiches and egg salad sandwiches. He will make no more than 20 sandwiches, because many people may bring their own. He figures that the number of chicken sandwiches should be at least twice as many as the number of egg salad sandwiches. Ken says that there is only enough chicken for 16 chicken sandwiches. He says that there should be at least 2 egg sandwiches..."\nI think these could be times in which you could use a system of inequalities. I guess if you had time to calculate it all...\n<333
247796
1 and 2/5 of a man or\n1 and 1/6 of the devil
357680
I'm trying 2 figure out how 2 chick this question.
189924
just add\n\n$02E2\n+ $85\n-----\nyou tell me...\njust remember to carry the one after hitting 14 (or E)
631878
If you are only concerned about the ways that you can place the 15 women between the 10 men, then you have 9 positions to place the 15 women.  (You don't have to worry about the arrangements for the men since there has to be one in between there is only one way to arrange.) These are the ways to do it (regardless of order).\n\na) 1 1 1 1 1 1 1 1 7\nb) 1 1 1 1 1 1 1 2 6\nc) 1 1 1 1 1 1 1 3 5\nd) 1 1 1 1 1 1 1 4 4\ne) 1 1 1 1 1 1 2 2 5\nf) 1 1 1 1 1 1 2 3 4\ng) 1 1 1 1 1 1 3 3 3\nh) 1 1 1 1 1 2 2 2 4\ni) 1 1 1 1 1 2 2 3 3\nj) 1 1 1 1 2 2 2 2 3\nk) 1 1 1 2 2 2 2 2 2\n\nNow to rearrange these, you would have to use the multinomial coefficient.  You have to use this because you have repeats: For example in a), you can switch two 1's around and get the same result, but moving the 7 will give you a different result.\n\nFor a), there are 9!/8!1! = 9 ways to arrange.\nFor b) and c), there are 9!/7!1!1! = 72 ways to arrange.\nFor d), 9!/7!2! = 36 ways.\nFor e), 9!/6!2!1! = 252 ways.\nFor f), 9!/6!1!1!1! = 504 ways.\nFor g), 9!/6!3! = 84 ways.\nFor h), 9!/5!3!1! = 504 ways.\nFor i), 9!/5!2!2! = 756 ways.\nFor j), 9!/4!4!1! = 630 ways.\nFor k), 9!/3!6! = 84 ways.\n\nSo in total, there are 9 + 72 + 72 + 36 + 252 + 504 + 504 + 84 + 756 + 630 + 84 = 3003 ways to place the 15 women between the 10 men.\n\nNow if after that you want to know how many ways there are to arrange if you do make a difference between men and between women, then multiply by the number of ways to arrange the men and women.\n\n3003*10!*15! = 14,250,102,106,034,995,200,000 ways.\n\nNotice this is less than the number of ways to arrange all 25 people regardless of gender.  That would have been 25! = 15,511,210,043,330,985,984,000,000.
562890
The flaw is in adding that 20.\nEach boy spent 90.\n90*3 = 270 of which 20 was pocketed by the son.\nIf you want how 300 was distributed then think like this\nActual cost was Cost of pen + 20 pocketed by son =270\nOf the original 300 only 30 was returned therefore 270+30=300\nThe point is either you do per boy calculation or for the whole group don't mix them and if you do mix then be cautious.\nFinally we conclude\n90+90+90 is the actual sum spent by boys whcih includes that 20 so don't add it again.
92903
Mr.\n\n===========\n\nThanks for the free 2 points.
619175
Assuming the division you're talking about is an infantry division, and omitting non-combat units, this is what I found:\n\nRegular Squad--12 soldiers assuming it's a rifle squad\nPlatoon--3 1/2 Squads (mortar squads have 6 men)\nCompany--4 Platoons (including Company HQ)\nBattalion--4 Companies (including Battalion HQ)\nBrigade--3 Battalions\nDivision--3 to 7 Brigades\n\nMy computation is from 10,000 to 14,112 fighting men. If I had included transport battalions, chemical companies, maintenance companies, support and detachment battalions, band companies, and other support and maintenance units I'd say about 18,000 to 20,000 men.\n\nOf course, armored divisions, field artillery batteries, air wings, and cavalry squadrons all have varying numbers.
1183230
The Laws of the Game do not state you have to start with 11 men. They state a maximum of 11 is allowed.\n\nTeams don't want to start with 10 because to start a game, the more the better. One theory would be that if you are then reduced to 10 men (or women!), you know you are up against it, put in more effort, which then actually over compensates for the missing man (or woman!).
813078
P(2 are women) = \nP(1st is woman) * P(2nd is woman) = \n6/10 * 5/9 = 30/90 = 0.333\n\nP(at least 1 is woman) =\n1 - P(both are men) =\n1 - P(1st is man)*P(2nd is man) = \n1 - (4/10)*(3/9) =\n1 - 12/90 =\n78/90 = 0.867\n\nI'm a little rusty on my "given that..." types of problems, and I'd hate to possibly steer you wrong on your remaining two questions.  But let me just say that I *THINK* the P of your 3rd question is 5/9; and I think the P of the 4th question is the same as the P of your 3rd question.\n\n~ ♥ ~
374315
options #6\n\n15,000 nukes in one's arsenal
512935
(a)\nC = mx + b\nC = 10x + 40\n(where fixed cost could be due to the cost of the workers or rental of the factory...)\n(The variable cost is the cost required to make 1 item)\n\n(b)\nR = 50x\n(The revenue, money collected, from selling x items is 50x.\nSo if you were to buy 1 item, you would pay $50\nand if you were to buy 3 items, you would pay $150.)\n\nRevenue = Cost\nR = C\n50x = 10x + 40\n50x - 10x = 40\n40x = 40\nx = 40/40\nx = 1\n\nThe selling of 1 item would result in the break-even point.
247553
Total no of items\n = 7 forks + 4 spoons + 5 knives \n = 16  \n\nIn the experiment of drawing 3 items from among 16\n\nTotal no. of possible choices \n= no. of ways in which the 3 items can be drawn from the total 16\n==> n = C(16,3)\n==> n = 16 * 15 * 14/3 * 2 * 1\n==> n = 560 \n[On the assumption that the drawing is being made without replacement]\n\nLet \n"A", "B", "C" and "D" be the events of the three items drawn being 3 forks, 2 forks and a knife, 3 spoons and 1 fork, 1 knife and 1 spoon respectively\n\nFor Event "A"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         3           0        0          3\nChoices       C(7,3)      C(5,0)    C(4,0)     C(16,3)\n\nNo. of favorable choices\n= No. of ways in which the 3 forks can be drawn from the total 7\n==> m(a) = C(7,3)\n==> m(a) = 7 * 6 * 5/3 * 2 * 1\n==> m(a) = 35\n\nProbability that the three items drawn are forks = m(a)/n\n==> P(A) = 35/560\n==> P(A) = 1/16\n\n\nFor Event "D"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         1           1        1          3\nChoices       C(7,1)      C(5,1)    C(4,1)     C(16,3)\n\nNo. of favorable choices\n= (No. of ways in which 1 fork can be drawn from the total 7)* (No. of ways in which 1 knife can be drawn from the total 5) * (No. of ways in which 1 spoon can be drawn from the total 4)\n==> m(d) = C(7,1) * C(5,1) * C(4,1)\n==> m(d) = 7 * 5 * 4\n==> m(d) = 140\n\nProbabilty for the 3 items drawn to be 1 fork, 1 kinfe and 1 spoon = m(d)/n\n==> P(D) = 140/560\n==> P(D) = 1/4\n\nYou should be able to solve the two parts in between for Events "B" and "C".\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
714317
problem number 1\nlet B = brian and M = maya. to find out when they catch each other we need to make the equation for brian's distance equal to maya's because that is when they will catch up:\n8 mph (x+1) = 10mph (x)   [remember x+1 is how long brian's been riding cuz he's been riding for one extra our than maya]\n8x+8 = 10x\n2x = 8\nx = 4\nso it will take Maya 4 hours to catch up\n\n2. x = hot dogs y = hamburger\n2x+1y = 3.99  (this shows 2 hot dogs multiplied by its price plus 1 hamburger multiplied by its price is 3.99)\n3x+3y = 6.57\n\nso since we need to find the cost of one hot dog (x) we need to make one of the two above equations into all x's.  so using the first one it would be y = 3.99-2x.  no using this equation, let's replace Y for the second equation.  \n3x+3y = 6.57\n3x+3(3.99-2x) = 6.57\n3x+11.97-6x=6.57\n11.97-3x=6.57\n11.97-6.57=3x\nx=1.80\nso one hot dog costs $1.80