Ask question

Ask question

All categories

2 years ago

13

A travel website sells tickets to a city for two airlines. Among those buying tickets, 20% fly airline A, and 80% fly airline B.

If the price of a ticket is $600 for airline A and $360 for airline B, what is the expected revenue of an airline ticket sold by this travel website? (Note: Airline A has much more desirable departure times than Airline B, so it charges more.)

1 answer:

Naddika [18.5K]2 years ago

7 0

Answer:

The expected revenue of an airline ticket sold by this travel website is $408

Step-by-step explanation:

For the revenues per ticket we have;

airline A; 600

airline B; 360

For the probability of choosing the airlines we have;

airline A; 20% = 0.2

airline B; 80% = 0.8

Therefore, the expected revenue of an airline ticket sold by this travel website is;

600(0.2) + 360(0.8) = 408

Therefore, the expected revenue of an airline ticket sold by this travel website is $408

You might be interested in

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

I need help on this one guys :(

Sophie [7]

Answer:

the last one

Step-by-step explanation:

6 0

2 years ago

17. Ennis has a length of rope that measures 3 3/4 yards. He cuts all of the

UNO [17]

Answer:

3 cuts

Step-by-step explanation:

if you take 3 3/4 -3/4 =3 2 cuts would give 3 yard size strips but you need to cut the 3/4 off so there would be 3 cuts for four pieces of rope

8 0

2 years ago

Which angle measures would make a triangle with three equal side lengths

Verizon [17]

Answer:

60

Step-by-step explanation:

If each side length must be identical, the same must be true for angles. If all the angles added are 180, divide 180 by 3 to get 60.

4 0

3 years ago

Read 2 more answers

Please answer its do in 2 hours

evablogger [386]

Answer:

both are reciprocals

Step-by-step explanation:

15.

$\frac{5}{8}\cdot \frac{16}{20}\\\\=\frac{5}{8}\cdot\frac{8}{5}=1$

so they are reciprocals

17.

$\frac{2}{3}\cdot\frac{12}{18}\\\\=\frac{2}{3}\cdot \frac{3}{2}=1$

so they are reciprocals

3 0

2 years ago