Draw a 4X4 grid. Draw an extra square on the lower left corner. (17 squares in all) In the upper right square draw a 'P' for prisoner. In the extra square draw an 'E' for exit. In the other 15 squares draw a 'G' for guard. In order to escape, the prisoner must go into each room with a guard and kill him. He must kill all the guards and cannot enter a room where a guard was killed. How does he escape?
217451
I think there are several answers.  You said he cannot enter a room where a guard was killed, but he CAN reenter the room he started in. \nSo, if you number the rooms like this:\n\n ,1  ,  2   , 3   , 4\n ,5   , 6   , 7 ,   8\n ,9    ,10   11   12\n13   14   15   16\n17\n\nAnd he started in 4, one answer would be if he went in this order:\n\nStarting in 4, then -3-4-8-7-6-2-1-5-9-10-11-12-16-15-14-13 and, finally-17.
372043
1. 101\n2.  There are 2 ones.  The first and last entries of the row.\n3.  Each entry is the sum of the 2 entries above and to     each side of it.\n4.  If the 2'nd entry in a row is a prime, it divides each of the other entries in the row (except for the last entry in the row which is always a 1)\n\nCheck out http://ptri1.tripod.com/ for all manner of kewl sh|t about Pascal's Triangle.  Also Fibonacci Series and other fun things ☺\n\n\nDoug\n\nEDIT:  Give Peter D the points.  He needs 'em worse than I do ☺
294273
11\nr,b.g.o\nr b g y\no y \n the next marble will make a set no matter the color\nyou could get a set before this if you draw well\nbut to be sure to get a set you need to draw 11
457468
Word problems like this can be hard.  Everybody has problems with them sometimes.\n\nIt does help to draw it out.  So you draw a little rectangle like this:\n ______________\n|                            |\n|                            |\n|______________|\n\n2 of the sides are 120 m and the other 2 are 150 m.  So you take 2 times each of these numbers and add them together:\n2(120) + 2(150)\n\nThat's your perimeter, in meters.  Now if your poles are 2 m apart, you don't need one for every meter you get in your answer, you only need 1 every 2 meters.  So you divide your answer by 2, and that gets your number of poles you need to put around the field.
139902
You have to graph them and find the corner points.  One of the corner points will be the answer.\nI found the corner points to be\n(28,52) and (40,40)\nPlug each point into g = 3x+y \nFor (28,52): g = 3(28) + 52 = 136\nFor (40,40): g = 3(40) + 40 = 160\n136 is the smaller value, so (28,52) is the answer.
408692
a) 24 in.\n\nSince a square has four sides...  You must add six for each side (i.e. 6 + 6 + 6 + 6) to get perimeter.
153517
This is a simple problem, so a straight-forward solution is to draw the set of all possibilities on a piece of paper and count them.  (And you should read this carefully, because it is the first answer so far to get all  parts correct.)\n\nProblem A)\n\nThere are 36 ways to roll 2 die:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 41\n36 35 34 33 32 31\n26 25 24 23 22 21\n16 15 14 13 12 11\n\nOf these, there are 11 ways to roll at least one 5 (Event A):\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ $$\n$$ 35 $$ $$ $$ $$\n$$ 25 $$ $$ $$ $$\n$$ 15 $$ $$ $$ $$\n\nSo, P(A) = 11/36\n\n\nProblem B)\n\nOf the 36 ways to roll 2 die, there are 26 ways for Event B to occur:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 $$\n36 35 34 33 $$ $$\n26 25 24 $$ $$ $$\n16 15 $$ $$ $$ $$\n\nSo, P(B) = 26/36 = 13/18\n\n\nProblem C)\n\nNotice that if Event A occurs, so does Event B.  Therefore, there are 11 ways for Event (A and B) to occur:\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ $$\n$$ 35 $$ $$ $$ $$\n$$ 25 $$ $$ $$ $$\n$$ 15 $$ $$ $$ $$\n\nSo, P(A and B) = P(A) = 11/36\n\n\nProblem D)\n\nNotice that if Event B does not occur, neither does Event B.  Therefore, there are 26 ways for Event (A or B) to occur:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 $$\n36 35 34 33 $$ $$\n26 25 24 $$ $$ $$\n16 15 $$ $$ $$ $$\n\nSo, P(A or B) = P(B) = 13/18\n\n\nProblem E)\n\nP(A|B) means the probability that Event A occurs given that Event B also occurs.  In this instance, there are only 26 ways to roll the dice so that B occurs:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 \n36 35 34 33 \n26 25 24 \n16 15 \n\nOf these 26 possibilities, there are 11 ways that Event A could occur:\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ \n$$ 35 $$ $$ \n$$ 25 $$ \n$$ 15 \n\nP(A|B) = 11/26\n\nAlternatively, you can directly use the definition of conditional probability, which is:\n\nP(A|B) = P(A and B)/P(B)\n\nSo, P(A|B) = (11/36)/(26/36) = 11/26
653935
what i did because to not fight that thing was escape before it does any damage the first one is to the left of the starting areathe next 3 are under electrical fields so be careful after stepping on the three plates go up the path on the right then do the same thing in the 3rd barrier the drop off theright side of the ledgeto a lower path then step on the platethen run up the slope and step on the highest plate before playing the keyboard  the next part is hard  if you need help i.m me at x_yunalesca_X93
315572
The best method to solve these type of problems is to sove them through VENN Diagrams. First draw a rectangle which is the sample space, then draw three circles intersecting each other wjich are E, F AND G. The common region between E & F is P[E n F] and so on. It is quite difficult to explain this in writing but I hope you have understood. Now by drawing Venn diagram I have tried to solve ur questions.\n\nPr[EuF'] = P[E] + P[F'] - P[E n F']\n         = 0.35 + (1 - 0.5) - (0.35 - 0.1)\n         = 0.6\n\nPr[F'nG']= 1 - P[F u G]\n         = 1 - ( P[F] + P[G] - P[F n G])\n         = 1 - ( 0.5 + 0.45 - 0.3)\n         = 0.35\n\nPr[E'nG] = P[G] - P[E n G] = 0.45 - 0.2 = 0.25\n\nI think above answers are correct and hope they help you!!
432041
I hope you meant to type "what is the width of the rectangle."  If there really is a triangle in this problem I'll need more info.\n\nAssuming it was the rectangle you meant to ask about, the first thing to do with a word problem is write equations for the data.  We'll call the width W, the length L and the perimeter P.  We're told that one side is 15 cm longer than the other.  We aren't told which is which, so I'll say it's longer than it is wide.  Therefore L = W + 15.  We're also told that the perimeter is 94 cm.  The perimeter is the sum of the lengths of all four sides, so it's 2L + 2W.\n\nSo our equations are:\nL = W + 15\n2L + 2W = 94.\n\nNow, substitute the first equation into the second:  2(W + 15) + 2W = 94.\n\nDistribute the 2:  2W + 30 + 2W = 94.\n\nSubtract 30 from each side: 2W + 2W = 64.\n\nCombine the Ws: 4W = 64\n\nDivide by 4: W = 16.\n\nNow go back to the first equation:  L = W + 15.  So L = 16 + 15 = 31.\n\nThe rectangle is 16x31 cm.  I don't know which is the length and which is the width, I just arbitrarily picked, so the answer could be either 16 or 31.  Hopefully you have additional information that will allow you to decide which it is.
528050
I suppose that it means all the lines that you can draw from every vertex to every other vertex.  For a rectangle it would be 2.  For a 5 sided figure.( I got up and sketched it) it would b 5.  You only count a line once, not both directions.\n\nAnother sketch suggests 9 for the hexagon.  Why don't you sketch a hexagon, draw and count the lines between vertices.
68519
Hello Michelle. Firstly, the word is Sudoku. I agree with Mayflyish in that the information given does not seem to work. For a 4x4 puzzle you need four numbers, letters or symbols. Also there is no single middle box. Additionally, using the layout you suggest (i.e. triangles and squares in each corner) means that none of the inner four boxes could contain a circle, because all of the outer vertical and horizontal lines between the squares and triangles would have to contain a circle each. So, I think, you will find that the four inner boxes must contain triangles and squares.
461519
just divide the inside by a perfect square (i.e. 4,9,16) and write it like this\n(this is the first problem you have)\nV(3X9) because 3X9=27\ntake the square root of the perfect square and move it to the outside\n3V3\nwhen you already have a number outside the radical, multiply it by the number you just simplified from the inside of the radical\n3V32\n3V(16X2)\n3X4V2\n12V2
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Yes, this sounds like the knight's tour.\n\nHere's a way to address it, but it may take fairly big sheet of paper.\n\nIn the corner, draw a circle inside of it place A1. This stands for the square in Row A, Column 1. Next to it draw two more circle, and label them B3 and C2. Draw a line from A1 to each of these. Draw all 30 squares as circles with collecting lines. It will look kind of like a map of cities connected by roads. Next. Start at A1 and visit each city once. It will be easier to see this pattern than to imagine the knights.\n\nOnce you're able to see the tour on your map, place the numbers back into the original table as you would have visited them.
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you divide the number or C.O.'s by the number of prisoners.\n\n18522 / 36620 = 0.50579   roughly  1 CO for every 2 prisoners
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ok so 2 length sides (6x-2) and 2 width sides(x+3), all added together, make the perimeter (100):\n2(6x - 2) + 2(x+3) = 100\nDistribute the 2s through the parenthesis:\n(12x - 4) + (2x + 6) = 100\nCombine like terms:\n14x + 2 = 100\nSubtract 2 from each side:\n14x = 98\nDivide each side by 14:\nx = 7
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Well, the second question is kinda fun... the fifth term has to be 2, so that the fourth term + the fifth term = the sixth term.  That means the seventh term is 2 + 6 = 8, and the eighth term is 6 + 8 + 14.\n\n* * * * \n\nOK, on the first one: Make a diagram so that you can visualize this.  You should see that the area of the square is the same as two right triangles, while the area of the trapezoid is the same as three right triangles.  So, the area of the trapezoid is 3/2 the area of the square, or 3/2 * 2800 = 4200.