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vivado [14]
3 years ago
13

There are 250 economy class seats on board. The airline sells some economy class seats at a discount price of $440 in advance an

d reserve a number of seats to sell at the full price of $770. Suppose that Dallas One Airline can always sell all discount tickets, and the demand for full price tickets is normally distributed with a mean of 100 and a standard deviation of 25. What is the probability that there will be no empty seat when the Dallas One Airline uses the optimal protection level? Choose the closest number.
Mathematics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

Check the explanation

Step-by-step explanation:

Cost of over reserving full cost tickets, Co = $440 ( empty seats as we cannot get even the discounted revenue)

Cost of under reserving Cu = 770-440 = $330 ( because you sell more discounted tickets)

Now critical ratio = Cu / (Cu+Co) = 330/770 = 0.42857

Now for above critical ratio ,z = -0.18 ( refer standard normal distribution table)

Optimal protection level = mean +z*standard deviation = 100-0.18*25 = 95.5 = 96

The possible costs include: all the above. You may need to provide compensation to passenger and make alternative travel arrangements. Also passenger won't be happy by over booking and their will be loss of good will.

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Answer:

Step-by-step explanation:

1) A perfect square is a whole number which is a product of a smaller whole number and itself. Examples of perfect squares are

4(2 × 2)

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25(5 × 5)

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2) Square root of 4x² is 2x(product of square root of 4 and square root of x²)

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The general formula for solving quadratic equations is expressed as

x = [- b ± √(b² - 4ac)]/2a

From the equation given,

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x = [- 20 ± √(400 - 400)]/8

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Step-by-step explanation:

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The density of a American white oak tree is 752 kilograms per cubic meter. If the trunk of an American white oak tree has a circ
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Check the picture below.

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\ \cline{1-1} C=4.5 \end{cases}\implies 4.5=2\pi r\implies \cfrac{4.5}{2\pi }=r \\\\[-0.35em] ~\dotfill

\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} h=8\\ r=\frac{4.5}{2\pi } \end{cases}\implies V=\pi \left( \cfrac{4.5}{2\pi } \right)^2(8)\implies V=\begin{matrix} \pi \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \cdot \cfrac{20.25}{\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}\underset{\pi }{\begin{matrix} \pi^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} }}(\stackrel{2}{\begin{matrix} 8 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}})

\bf V=\cfrac{40.5}{\pi }~cm^3~\hspace{9em} \stackrel{\textit{since density is }752kg~per~cm^3}{density\implies 752\left( \cfrac{40.5}{\pi } \right)}\qquad \approx ~~9694

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3 years ago
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