The county hospital is located at the center of a square whose sides are 3 miles wide. If\nan accident occurs within this square, then the hospital sends out an ambulance. The\nroad network is rectangular, so the travel distance from the hospital, whose coordinates\nare (0, 0) to the point (x, y) is |x| + |y|. If an accident occurs at a point that is uniformly\ndistributed in the square, find the expected travel distance of the ambulance.
136810
I assume you know integration.\n\nBreak the square into 4 parts. Take the positive quadrant. (1.5m x 1.5m). Now to reach (x,y) we have to travel x+y.\n\nExpected distance = Sum (distance of point * p(accident at that point) )\nHere p (accident at point) = 1/1.5^2 = 1/2.25\n\nExp distance = sum(distance of point) / 2.25\n\nIntegrate (x+y) with respect to x and y from 0 to 1.5 ( [ stands for integration)\n\n [ [ (x + y) dx dy\n   = [ ( x^2/2 + xy/2 ) dy\n   = [ (2.25/2 + 1.5y/2) dy (x ranges from 0 to 1.5)\n   = 2.25y/2 + 1.5y^2/2\n   = 3.375/2 + 3.375/2   (y ranges from 0 to 1.5)\n   = 3.375\n\nExp distance = 3.375/2.25 = 1.5\n\nSince all quadrants are similar. The expected travel distance is going to be 1.5 miles.\n\nAfter solving this, I got a simple solution. Seperate the event into 2 events. One in x direction and another in y direction.\n\nExp distance in x direction = 1.5/2 = 0.75\nExp distance in y direction = 1.5/2 = 0.75\nTotal exp distance = 1.5 (they are independent events)
486546
A.\nThey drove more distance on the highway than in the city, which come from 2 reasons:\n\n1. They spent more time on the highway, viz. 50 min.  at 60mph this would be 50 miles.\n2. Their average speed on the highway was 60mph, and only 30mph in the city. For equal times, the distance difference will be at a rate of.60-30 = 30mph\nThe total difference between highway and city distances is 100 miles.  Of this, 50 miles are due to extra time on the highway.  So 50 miles are due to speed difference.  The time spent in the city is, therefore, 50/30 hrs = 100 mins.\n\nSo they spent 100mins in the city and 100+50 = 150 mins on the highway.  So total distance=\n30*100/60 + 60*150/60\n= 50 + 150\n=200 miles.\n\nB.\nAnother way. Let the distance travel led in city be x miles.  Distance traveled on highway = x + 100 miles.\ntime in city = x/30 hrs\ntime on highway = (x+100)/60 hrs.\n\ntime difference:\n(x+100)/60 - x/30 hrs\n= x + 100 - 2x mins\n=100 -x mins.\n\nbut we know this is 50. so\n100 - x = 50\n>>\nx = 100 - 50 = 50. (=distance in city)\n\ndistance on highway = 100 + x = 150.\n\nTotal distance = 50 + 150 = 200 miles.
137884
Let t be the time it takes for Denise to catch Gloria expressed in hours.\n\nGloria will have been walking for (t + 1/2) hours, at 3 mph\nso she will have gone (t+1/2) * 3 miles.\n\nDenise will have been jogging t hours at 5 mph, so she will have gone 5t milse.\n\nSince they are at the same place, the distances will have to be the same: 3(t+1/2) = 5t\n\nSolve for t:\n3t + 3/2 = 5t\n\n3/2 = 2t\n\n3/4t = t\n\nCheck Denise goes 3/4 * 5 = 15/4 miles\n\nGloria goes (3/4 + 1/2) * 3 = (5/4) * 3 = 15/4 miles\n\n\n\nIf you have a square, and you triple the length of the side,  the resulting square has an area that is 9 times that of the original square.  The same thing works for all figures in the plane. If you enlarge them by a factor of 3, the area grows by a factor of 9.   \n\nIf you were to break the star down into a collection of triangles, and analyzed the area of each one, you would see that you were multiplying the base and altitude of each triangle by a factor of three, so you are multiplying the area of every one of the pieces by 9. So 90 is your correct answer.
184248
A car is traveling at a rate of 100 kilometers per hour to a city that is 350 kilometers away. Write an equation that represents the remaining distance to the city at any given time of travel.\n\ny = remaining distance\nx = time\n350 is the distance away\n-100 is the distance closed per hour\n\ny = -100x + 350
289366
Contact the hospitals in Green River. It sounds fairly small, and ER is bound to remember a train/pedestrian accident. You can look up the phone numbers to all the hospitals in the town on smartpages.com.
258112
OK, I drew the picture and this is what I came up with.\n\nThe anchor point or bottom of the bridge touches the ground at the edge of the shoulder.  Lets call the middle of the road on the ground (0,0).  So, the anchor point is at (8,0).\n\nIt also says that when you get to the edge of the shoulder and the road that the clearance must be 2.5 meters.  This point would be (6,2.5).  Symmetry gives us (-8,0) and (-6,2.5).  And just for fun, note that the parabola is open down, just in case that comes up.\n\nThe generic quadratic is ax^2 + bx + c.\n\nLets put some of our points in and solve for a,b, and c.\n\na(8)^2 + b(8) + c = 0.  So, c = -64a - 8b  using (8,0)\n\na(-8)^2 + b(-8) +c =0.  So, c = -64a + 8b using (-8,0)\n\nSetting these 2 values for c equal to each other I get that b=0. (We should have known this already because we put the vertex directly on the y-axis.  This will always give you a b value of zero because the quadratic has to be able to factor into the form \n(sqrt(a) - sqrt(c))(sqrt(a) + sqrt(c))\n)\n\nSo, y = ax^2 + c.  Lets put a third point in: (6,2.5)\n\na*36 + c = 2.5 or c = 2.5 - 36a.  From directly above we have that c = -64a - 8b, but b turned out to be 0, so,\nc = 2.5 - 36a = -64a  --> 2.5 = -28a  \n--> a = -2.5/28 = -5/56\n\nc = 2.5 - 36(-5/56) = 2.5 + 45/19  = 95/19 + 45/19 = 140/19.\n\nI come up with y = -5/56*x^2 + 140/19 as the equation for the bridge.  The highest point will be when x = 0, so, the maximum clearance is at (0,140/19).
469570
Let D = distance between home and school.\n\nUsing the formula : distance = speed * time, we have :\n\nFrom home to school : D = x * T1\nFrom school to home : D = kx * T2\n\nTherefore, x * T1 = kx * T2\nDividing through by x gives :\nT1 = k * T2\n\nWe know that T1 + T2 = 1 (total time travelled)\nTherefore, T2 = 1 - T1.\n\nSubstituting this into T1 = k * T2 gives :\nT1 = k (1 - T1)\n\nFrom this, we find that :\nT1 = k / (k + 1)\n\nSubstituting into 'home to school equation' gives :\nD = x * k / (k + 1)\n\nTherefore total distance covered\n= 2D\n= 2 * x * k / (k + 1) miles.
113940
The way I draw the picture is the port is at the origin.  Then there is a line in the 2nd quadrant 53 degrees up from the x axis, and another in the 3rd quadrant 67 degrees down from the x axis.  This gives the total angle between the 2 as 120 degrees.  If it's supposed to go the other way, 53 degrees left of the y-axis and 67 degrees left of the y-axis, then the total angle between the 2 would be 60 degrees.\n\nEither way, you use law of cosines to solve.  You have the north side of the triangle at 36 mi (3hr * 12 mph) and the south side at 48 mi (3hr * 16 mph) with I'm going to use the 120 degree angle between.\n\nLaw of cosines if you have 2 sides of a triangle, and the angle between them is\n\nc^2 = a^2 + b^2 - 2ab*cos(C) - we'll let a = the 36 side, b = the 48 side, and c is the distance between the ships at noon.\n\nc^2 = 36^2 + 48^2 - 2(36)(48)(cos 120)\nc^2 = 1296 + 2304 - (3456)(-1/2)\nc^2 = 5,328\nc = approx 73 miles.\n\nIf you think it should be a different angle, just re-solve the problem with that angle instead, but this is how you would do it.
301879
crackman's answer is right...he used the cosine law to get the third distance..\n\ndistance^2=85^2+102^2-2(85)(102)(cos85)\nget rid of the square by getting the square root of the right side of the equation....that's how he got it...
548825
Put on a number plane with B initially at O (0. 0)\nTherefore A is initially at (-150, 0)\n\nNow A's position, t hours after midday, is (-150 + 35t, 0) and B's position is (0, 25t)\n\nSo t hours after midday \n\n|AB| = √[((-150 + 35t) - 0)² + (0 - 25t)²]\n= √[(22500 - 10500t + 1225t² + 625t²)\n= √(22500 - 10500t + 1850t²) \n\nVelocity of change = d(|AB|)/dt\n= 0.5 * (3700t - 10500) / √(22500 - 10500t + 1850t²)\n= 0.5 * (3700 * 4 - 10500) / √(22500 - 10500 * 4 + 1850 * 4²)\n= 0.5 * 4300 / √(10100)\n≈ 21.4 km/h
410865
At 1 p.m the 1st bus starts moving west with a velocity of 44 mph  that means at 2 pm the bus will be 44 miles away from the station because  D = V x t = 44 mph x 1 h = 44 miles .\nAt 2 pm the 2nd bus starts moving east i.e in the opposite direction with a velocity of 48 mph \n              \nB1  ____________B2      ( B1 B2 = 44 miles )\n                      \n             \n\nAfter 2 pm the distance between the 2 buses at any time is equal \nto : \nD = 44 +( V1 x t ) + (V2 x t ) = 44 + (44 x t ) + (48 x t ) = 44 + 92 t \nwhere D >> miles , t >> hrs \n                                         \n                                          B1    _________S___________ B2\nWhere B1 S = 44 miles + 44 t \n            B2 S = 48 t \n                                             \nPut D = 274 miles to get  t \n274 = 44 + 92 t     >>  t = 2.5 h\nTherefore the two buses will be a distance of 274 miles apart at : \n2 pm + 2.5 h =  4 : 30 pm
340304
1st way :-\ndistance = speed x time = 55 mph x 3 h = 165 m\n\n2nd way :-\ngiven that d = 55t , so on putting t = 3hrs , we get d = 165 m
451757
d= v .t\n> d = 50 km/hr (18 min)\n>d = 50 km/60 min (18min)\nd= 50 km (18)/60\nd =900 km/60\nd = 15 km >>the distance travelled\n\nIf you walk at an average of 3 miles per hour (approximately 4.8 km/hr) at a distance of 15 km:\nt= d/v\n=15 km/4.8 km/hr\nt = 3.125 hrs or (3 hours and 7.5 minutes)>>the time you need to walk on 15 km at the speed of 4.8 km/hr.♥
299433
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
142473
Let x = Eloise's time\nChristian's time = x-3\nsince Chistian's time is three hours less than Eloise's time\nNow distance equals rate times time or\nd = r*t\nFor Eloise r = 15, so d = 15x\nFor Christian, r = 30, so d = 30(x-3)\nThese distances are equal so\n15x = 30(x-3) \nDistribute 30\n15x = 30x - 90  \nSubtract 30x from each side\n-15x = -90 \nDivide both sides by -15\nx = 6\nThus, they traveled\nd = 15x\nd = 15(6)\n= 90 miles
196125
Let's think about how to solve these and maybe that will help you with others,\n\n1. Christian makes the trip twice as fast as Eloise.  Therefore his trip would take half the time.  If his trip was three hours shorter, then 3 hours is half the time.  His trip was 3 hours, Eloises was 6 hours.  So if Christian traveled at 30 mph for 3 hours, and Eloise traveled at 15mph for 6 hours .. how far is that?\n\n[modified below with details for additional answers]\n2. Basically \n4mph x SomeTime = SomeDistance\n5mph x (Sometime-1 hour) = SomeDistance\n\nor, \nif 4x=y and 5(x-1)=y\nthen,\n4x=5x-5\n4x-5x=-5\n-x=-5\nx=5  (this is the SomeTime)\n\ngoing back to the first equation\n4mph x SomeTime = SomeDistance\n4mph x 5hours = 20 Miles\n\ndoublecheck with the second equation,\n5mph x (Sometime-1hour) = SomeDistance\n5mph x (5hours-1hour) = 5mph x 4hours = 20 miles\n\n\n3. If the fast freight was going 60mph for 6 hours (noon to 6pm), then the whole trip is how far? .. 360 miles.  If the slow freight takes 8 hours to go 360 miles, then what is the speed of the slow freight?  360mph/8m = 45 mph
596961
displacement = final position - initial position.\nBecause the path is a square, you need two coordinates (x, y) to specify position. Besides the start poit must be specified.\n\nIf we place the rectangle over the 4th quadrant, and the farmer starts walking from the corner at point (0, 0) and moving clockwise then he will move 140 m in 140 s (2 min 20 sec)(because he moves 40 m in 40 s, speed: 1 m/s)\n\nAt the end he will move around the square 3 times and a half, it means he is in the opposite corner where he started.\nThe displacement is the diagonal of the square, that you can calculate by Pithagoras' theorem:\nD = sqrt(10² + 10²) = 14.14 m\n\nIf he starts moving from another initial position, the displacement could be different.