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marishachu [46]
3 years ago
12

Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22. (Past studies have revealed that onl

y 86% of the booked passengers actually arrive for the flight.)Find the probability that if Air-USA books 24 persons, not enough seats will be available.prob =Is this probability low enough so that overbooking problems will be unusual? (Define unusual as having probability 5% or less)A. yes, it is low enoughB. no, it is not low enough
What about defining unusual as 10% or less?A. no, it is not low enoughB. yes, it is low enough
Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

B. no, it is not low enough

A. no, it is not low enough

Step-by-step explanation:

Given that Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22.

Prob for  a random person booked arrive for flight = 0.86

No of persons who books and arrive for flight, X is binomial, since there are two outcomes and each person is independent of the other

The probability that if Air-USA books 24 persons, not enough seats will be available

= P(X=23)+P(x=24)

= 0.1315

B. no, it is not low enough

-------------------------------

The prob we got is >10% also

A. no, it is not low enough

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For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim
Marina86 [1]

Area of the shaded region $=36(\pi -2) square cm

Perimeter of the shaded region =6 (\pi + 2\sqrt 2) cm

Solution:

Radius of the quarter of circle = 12 cm

Area of the shaded region = Area of quarter of circle – Area of the triangle

                                             $=\frac{1}{4} \pi r^2 - \frac{1}{2} bh

                                             $=\frac{1}{4} \pi \times 12^2 - \frac{1}{2} \times  12 \times 12

                                             $=36\pi -72

                                             $=36(\pi -2) square cm.

Area of the shaded region $=36(\pi -2) square cm

Using Pythagoras theorem,

AC^2=AB^2+BC^2

AC^2=12^2+12^2

AC^2=288

Taking square root on both sides of the equation, we get

AC= 12\sqrt 2 cm

Perimeter of the quadrant of a circle = \frac{1}{4} \times 2\pi r

                                                             $=\frac{1}{4} \times 2 \times \pi \times 12

                                                             $=6 \pi cm

Perimeter of the shaded region = 6 \pi + 12\sqrt 2 cm

                                                    =6 (\pi + 2\sqrt 2) cm

Hence area of the shaded region $=36(\pi -2) square cm

Perimeter of the shaded region =6 (\pi + 2\sqrt 2) cm

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3 years ago
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NEED HELP ASAP<br> Solve 4 – x = –8.
Anna71 [15]

Answer:

i think it is 12 positive

Step-by-step explanation:

i think

4 0
3 years ago
HELPme
aalyn [17]

Answer:

1. Rectangle

2. 376.8 cm

3. 379.94 square feet

4. –4 and 4

Step-by-step explanation:

A square is always a rectangle.

If a wheel has a radius of 15 cm, it would travel approximately 376.8 cm per 4 revolutions.

If the diameter of a circular garden is 22 feet, the approximate area of the garden is 379.94 square feet.

The solutions to the equation y2 – 1 = 15 is –4 and 4.

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