A car stopped 22.8 times according to calculation. Should the answer be 22 or 23 times?
822655
How is the problem phrased?  Most statistical questions are ok with a 22.8 answer.  Like in the US, most families have 2.4 kids... they don't really have .4 of a kid, but that it is an appropriate way to give a stat.  \n\nOtherwise, if you just need to know something like "how many times could the car have possibly stopped," then 22.\n\nIt would help to know what the problem says and what class and grade you're in. =)
201384
The correct answer is C, although it can vary slightly depending on the vehicle, suspension set-up, and braking system
119644
22,000 mph = 28.901466396 mach
903925
Total cost = 12.90 + (c*1.20)\n\nThe total number of cars is a bit off - the variable costs are $26, which leaves a remainder when you divide by 1.20. You could make 21 cars, but that should cost $38.10 (21x1.2 + 12.90). 22 cars would cost $39.30...
485266
Divide 20 by 5 to see how many 5 minute increments are in the trip, then multiply by 2 for the miles covered.\n\n20/5 = 4*2=8
715233
If it is an arithmetic sequence, then the number between should be exactly half way.\n\n214 and 225 are 11 apart, so you want to pick the number that is 5½ higher or lower (namely 219½).  The pattern should follow for all numbers earlier or later.\n\nSo the answer is:\n...208½, 214, 219½, 225, 230½...
926411
The stop distance is not an arithmetic progression (ie it is not 40m). It is in fact a geometric progression. The stop distance is a lot lot more. Truthfully, I don't know but if you were travelling at 120 kmh and there was a truck 40m (or even 60m) in front of you when the brakes locked; you'd be dead.
985663
First, if you are just going to have 3 numbers from 0 to 27 inclusively, you have 28 numbers all in all and it is with repitition so you will have just raise the 28 to the 3rd power \n\nso you will get 21,952   \n\nand then , if you will have a correct answer on the first time you will get \n\n1/21952
966205
(a) too trivial to answer\n(b) Take the average speed times the number of seconds over which it had that average speed.
1175600
They will meet when:\n\n4mph x Time = 3miles - 5mph x Time\n\nSo the answer is 0.33333 hours \n\nOr just 20 minutes.\n\n(not 27)
103191
There are three stops, 20 minutes a piece, giving us 1 hour. The plane travels at 200 MPH and needs to cover 3000 miles. 3000 divided by 200 is 15, 200 goes into 300 once, 300 - 200 is 100, bring down the zero and you get 1000, 1000 divided by 200 is 5, so you get 15 with no remainder. 15 hours plus 1 hour is 16 hours. However, if you don't count the landing as a place that takes 20 minutes, if you count the landing as the end, 20 minutes would be subtracted giving you 15 hours 40 minutes.
859010
The average veocity of the car from 0 to 25 m/s is 12.5 m/s.\n\nThe total amount of time it takes is 200 m / (12.5 m/s) = 16 seconds.\n\nThe change is speed in 16 seconds is from 0 to 25 m/s.  Therefore, the acceleration is (25 m/s)/ (16 s) = 1.5625 m/s^2\n\nThe force = mass * acceleration = 1600 kg * 1.5625 m/s^2 = 2500 N\n\nAnswer is d.
1134456
On a stretch of road that ran straight for about 3.5 miles, I clocked a oncoming vehicle at 136mph. I had never had one that high and was wondering if the radar was not accurate. I reset it and it showed 147mph, to my luck another unit was behind me about 1.5 miles and was able too stop the vehicle. Needless to say no ticket issued, he was jail bound and his vehicle towed. Not had one that high since.
600121
Your logic is flawed. Say you always pick card #2. The first time, you'll have 14.2857...% (that's the same as saying one seventh) of the chances of winning the car.\n\nThe second time, with a new set of cards, your chances that the card #2 is the winner are again 14.2857...%, and so will every new time, because your chances are always 1 out of 7. You are picking one card out of seven cards that you are given.\n\nYour reasoning would be right if you could be assured that the car card will always be a different one (that for example, it will be in the first case #4, then #6, and so on). In that case, after the 7th. guess (if not earlier), you'd get the car. But the way you explained the game, nothing prevents the car from being always for example at card #5. If that is the case, always trying card #2 will make you fail every time.
943121
Both of the top 2 answers are correct depending on what you meant. If you meant the sum of 5 times EACH of the numbers then the first answer is correct. If you meant the smaller number added to 5 times the larger number then the second answer is correct.
884244
a. 1.10 MJ\n\nEnergy = mass * acceleration * distance = m * a * s\n\nm = 1600 kg\nAssume that the car is accelerating at 1 m/s^2 =>  a = 1\n\nIt would take 25 seconds for a car accelerating at 1 m/s^2 to go from 15 m/s to 40 m/s.\n\nThe average velocity during the acceleration is (15 + 40)/2 = 27.5 m/s.\n\nThe distance, s, is 27.5 m/s * 25 seconds = 687.5 m\n\nm*s*a = 1600 kg * 1 m/s^2 * 687.5 m = 1.1 e6 J = 1.1 MJ\n\nIf you assume another acceleration (let's say 5 m/s^2) you will get the same because the same amount of work is being done.
786122
Acceleration is defined as the change in velocity (speed) with respect to time. In your problem statement, the phrase "slowing down uniformly" implies that the change in velocity is happening at a constant rate.\n\nKnowing this, you can compute the acceleleration as:\n\na = (v2 - v1) / t\n\na = (5 - 20 m/s) / (4 s)\n\na = -3.75 m/s²\n\nUNDERSTAND THIS: The negative sign means that the velocity is slowing down. It does not imply that the velocity is negative.\n\nFor part (a), force is defined as the product of mass and acceleration. In other words, a force of 1 N is required to acclerate a 1 kg mass at a rate of 1 m/s². Mathematically:\n\nF = m * a\n\nYou're given the mass of the car, and you now know its acceleration, so the average force acting on the car is simply:\n\nF = m * a\n\nF = (2000 kg) * (-3.75 m/s²)\n\nF = -7500 N\n\nThe negative sign means that the force is acting against the car's direction of travel.\n\nFor part (b), the distance the car travels in this time is calculated by the distance equation:\n\nS = (1/2) * a * t² + vi * t + Si\n\nWhere vi is the car's initial speed and Si is the car's initial displacement (distance). Initially, at t=0 s, we'll assume that the initial displacement is 0 (because we only want to know how far the car travels from when we start the clock at t=0). Additionally, vi is given as 20 m/s. Plugging everything into the distance equation provides:\n\nS = (1/2) * a * t² + vi * t + Si\n\nS = (1/2) * (-3.75 m/s²) * (4 s)² + (20 m/s) * (4 s) + 0\n\nS = -30 m + 80 m\n\nS = 50 m\n\nThe car travels 50 m during this period of slowing down.\n\nHope that makes it all clear! :)