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kodGreya [7K]
4 years ago
11

at maximum speed, an airplane travels 1720 miles against th wind in 5 hours. Flying with the wind, the plane can travel the same

distance in 4 hours. Let x be the maximum speed of the plane and y be the speed of the wind. What is the speed if the plane with no wind?
Mathematics
1 answer:
iragen [17]4 years ago
4 0

Answer:

x = 387

Step-by-step explanation:

The wind acts in the opposite direction to the plane when the time is longer.

The wind acts in the same direction as the plane when the time is shorter.

Equation

d = r*t

Givens

  • d = 1720
  • x = plane's speed
  • y = wind speed
  • t1 = 5 hours
  • t2 = 4 hours.

Substitution

  • d = (x + y) * t
  • 1720 = (x + y) * 4
  • 1720 = (x - y) * 5

Since the two distances are the same, you can equate the right hand sides.

  • (x + y) * 4 = (x - y)*5              Remove the brackets.
  • 4x + 4y = 5x - 5y                  Add 5y to both sides
  • 4x + 4y + 5y = 5x - 5y + 5y  Combine all the terms.
  • 4x + 9y = 5x                          Subtract 4x from both sides
  • 9y + 4x - 4x = 5x - 4x           Combine
  • 9y = x

Now use one of the given equations with the distance

  • 1720 = (x + y)*4           Substitute 9y for x
  • 1720 = (9y + y) *  4      Combine
  • 1720 = 10y * 4             Multiply the right
  • 1720 = 40y                  Divide by 40
  • 1720/40 = 40y/40      Do the division
  • 43 = y                          y = wind speek
  • 9y = x                          x = 9*y  
  • 9*43 = x
  • x = 387 miles per hour.
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The probability mass function for the Binomial distribution is given as:

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a. What is the probability that you count exactly 10 in poverty?

For this case we want this probability P(X=10)

P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928

b. What is the probability that you count 10 or less in poverty?  .2614

For this case we want this probability P(X=\leq10)

P(X\leq10)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)

And we can find the individual probabilities like this:

P(X=0)=(100C0)(0.127)^{0} (1-0.127)^{100-0}=1.263x10^{-6}

P(X=1)=(100C1)(0.127)^{1} (1-0.127)^{100-1}=1.837x10^{-5}

P(X=2)=(100C2)(0.127)^{2} (1-0.127)^{100-2}=0.000132

P(X=3)=(100C3)(0.127)^{3} (1-0.127)^{100-3}=0.000629

P(X=4)=(100C4)(0.127)^{4} (1-0.127)^{100-4}=0.00222

P(X=5)=(100C5)(0.127)^{5} (1-0.127)^{100-5}=0.00620

P(X=6)=(100C6)(0.127)^{6} (1-0.127)^{100-6}=0.0143

P(X=7)=(100C7)(0.127)^{7} (1-0.127)^{100-7}=0.0279

P(X=8)=(100C8)(0.127)^{8} (1-0.127)^{100-8}=0.0471

P(X=9)=(100C9)(0.127)^{9} (1-0.127)^{100-9}=0.0701

P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928

And then repplacing we got:

P(X \leq 10) = 0.2614

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For this case we need after 7 people , 1 in poverty so we can find this probability like this:

(1-0.127)^7 (0.127) =0.0491

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