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Maslowich
3 years ago
10

On a no wind day, an airplane is flying 2000 miles round trip from Chicago to Boston and back to Chicago with a constant speed o

f 500 mph. About how long would it take to fly the same round trip if there is a wind speed of 100 mph from Chicago to Boston during the whole round trip flight?
Mathematics
1 answer:
Greeley [361]3 years ago
3 0

Answer:

It would take the plane about 4.17 hours to fly the whole round trip.

Step-by-step explanation:

Assuming that the distance from Chicago to Boston is 2000/2 = 1000 miles.

Since the wind is blowing from Chicago to Boston, when the plane is traveling to Boston, the velocity of the wind adds to the its velocity:

v_1 = 500mph + 100mph = 600mph;

therefore, the time it takes for the plane to go from Chicago to Boston is

t_1 = \dfrac{1000miles}{600mph}

\boxed{ t_1 = 1.67hr}

Now, as the plane flies back from Boston to Chicago, the wind is blowing opposite the direction of its movement; therefore, the wind's velocity subtracts from the velocity of the plane:

v_2 = 500mph-100mph =400mph

and the time taken during the trip is

t_2 = \dfrac{1000miles}{400mph}

\boxed{t_2 = 2.5hr. }

Therefore, the total time taken t_{tot} during the whole round trip is

t_{tot}= t_1+t_2  \\\\t_{tot} = 1.67hr+2.5hr \\

\boxed{t_{tot} = 4.17hr}

Therefore, on the windy day, it would take the plane 4.17 hours to fly the whole round trip

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<span>3 + (-1/4)² ÷ 1/2
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The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that
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Answer:

a) P(X∩Y) = 0.2

b) P_1 = 0.16

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Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

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At the same way, the probability P_2 that he must stop at the second signal but not at the first one can be calculated as:

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P = P_1+P_2\\P=0.16+0.31\\P=0.47

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