What is the formula for an eastbound car is going 4 miles per hour faster than a westbound car. the cars are 208 miles apart 2 h
ours after passing each other on a highway. what is the speed, in miles per hour, of the eastbound car?
1 answer:
Draw a diagram to illustrate the problem as shown below.
Let v = speed of the westbound car, mph
Because the eastbound car travels 4 mph faster than the westbound car, its speed is (v+4) mph.
After 2 hours,
the westbound car travels 2v miles west, and the eastbound car travels 2(v+4) miles east.
Because they become separated by 208 miles, therefore
2v + 2(v+4) = 208
4v + 8 = 208
4v = 200
v = 50 mph
The westbound car travels at 50 mph.
The eastbound car travels at v+4 = 54 mph
Answer: The eastbound car travels at 54 mph.
Let v = speed of the westbound car, mph
Because the eastbound car travels 4 mph faster than the westbound car, its speed is (v+4) mph.
After 2 hours,
the westbound car travels 2v miles west, and the eastbound car travels 2(v+4) miles east.
Because they become separated by 208 miles, therefore
2v + 2(v+4) = 208
4v + 8 = 208
4v = 200
v = 50 mph
The westbound car travels at 50 mph.
The eastbound car travels at v+4 = 54 mph
Answer: The eastbound car travels at 54 mph.
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a)
b)
c)
With a frequency of 4
d)
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
Replacing we got:
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
Part c
The mode is the most repeated value in the sample and for this case is:
With a frequency of 4
Part d
The midrange for this case is defined as:
Part e
For this case we can calculate the deviation given by:
And replacing we got:
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case