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pochemuha
3 years ago
6

Juan is participating in a 40 mile bike race. He will pedal at a steady rate of 11.5 miles per hour. He pedaled for 1 hour and 4

5 minutes with the wind and for 2 hours and 30 minutes against the wind and finished the race in 4 hours and 15 minutes. What was rate at which the wind was blowing in miles per hour?
Mathematics
1 answer:
Archy [21]3 years ago
7 0

To work this problem, you must assume that Juan's pedaling speed adds to the wind speed when going downwind, and that the wind speed subtracts from Juan's pedaling speed going upwind. (In other words, Juan's pedaling speed is relative to the air, not the ground.)

Let w represent the wind speed in miles per hour.

... distance = speed × time

Juan's total distance is 40 miles, so we have ...

... 40 = (11.5 +w)×1.75 + (11.5 -w)×2.50

... 40 = 48.875 - 0.75w . . . . . simplify

... -8.875/-0.75 = w . . . . . . . . subtract 48.875, divide by -0.75

... w = 11.833... = 11 5/6

The wind was blowing 11 5/6 miles per hour.

_____

Compared with real-life experience, and working through the details, this problem makes no sense whatever. Pedaling a bicycle is not like rowing a boat, where the current of the medium directly affects speed.

If you work through the segments of the problem, you find that Juan traveled 40.833 miles with the wind in the first 1.75 hours, so actually finished the race and then some. He spent the next 2.5 hours being blown backward by the wind, even though he was pedaling forward at 11.5 mph. (How much sense does that make?)

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tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

<u><em>Step(ii):-</em></u>

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

<u><em>Final answer:-</em></u>

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

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3 years ago
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Answer:

<em>Observe attached image</em>

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x = 4

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