If S=( 1/5, 1/4, 1/3, 2/3, 3/4, 4/5) and x and y are elements of S, what is the maximum possible value of 3x-3y?
On a given day there are 12 trains to city X with an average of 1400 commuters per train. if the number of trains were cut to 7 and the total number of commuters remained the same, there would be an average of how many more commuters per train?
i need the steps please!! thanks thanks very muchhh………..
15 + 18 + 21 = 54, but since they are pairs, divide by 2, 54/2 = 27,and 27/3 = 9.
9 is the average for the three numbers.
The maximum possible value means taking the smallest number from the greatest, therefore, 4/5 is the greatest number, and 1/5 is the smallest.
3x – 3y = ? Inserting these numbers gives
3(4/5) – 3(1/5) = ? or 12/5 – 3/5 = 9/5
9/5 is the maximum possible value
12 trains @ 1400 commuters per train gives
12 x 1400 = 16,800 average total commuters
reduce the trains to 7 and you get
16,800/7 = 2400 average commuters per train
The average increase is 1000 commuters per train.
Add together to get 2x+2y+2z=54 so x+y+z=27 so average=27/3=9.
#2 max value for x-y is biggest-smallest
max value= 4/5-1/5=3/5
max for 3x-3y=3(3/5)=9/5
#3 number of commuters=12 x 1400
average per train =1400
with 7 trains, average per train= (12×1400)/7=2400
increase in average per train= 2400-1400=1000 passengers
2)max value = 3*(4/5) – 3*(1/5)
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