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raketka [301]
3 years ago
11

A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If

all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses
Mathematics
1 answer:
BartSMP [9]3 years ago
5 0

Answer:

Step-by-step explanation:

Let the speed of cyclist be C and bus be B.

Using the concept of relative speed:-

For incoming bus,speed = B + C

For overtaking bus,speed = B - C

Now,distance travelled b/w 2 incoming bus = (B + C)*4

Now,distance travelled b/w 2 overtaking bus = (B - C)*12

so,

(B + C)*4 = (B - C)*12

or, B=2C

Hence,distance b/w two buses in same direction,d=4*(3/2)B=6 B

Therefore, time gap=d/B=6 minutes

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The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s
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L= Q_1 - 1.5*IQR = 210 -1.5*40 =150

U= Q_3 + 1.5*IQR = 250 +1.5*40 =310

And if we analyze the info provided we have 500 values

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So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

IQR = Q_3 -Q_1 = 250-210 =40

For this case we need to find the lower and upper limit with the following formulas:

L= Q_1 - 1.5*IQR = 210 -1.5*40 =150

U= Q_3 + 1.5*IQR = 250 +1.5*40 =310

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

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Answer:

x=y/10-2.875

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Standard form; 40x-4y=-115

Step-by-step explanation:

Not sure which answer you're looking for, but hope this helps!

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