Ask question

Ask question

All categories

2 years ago

14

The administrator of a county's museum is considering increasing the price of an annual pass by 1% from $55 to $55.55. At the cu

rrent price, the museum sells 2000 annual passes and the point of elasticity is 0.62. What effect would the increase in price have on the museum's annual pass revenue?

1 answer:

atroni [7]2 years ago

8 0

The increase in price would increase the annual pass revenue of the museum.

<h3>What is the effect of the increase in price on the annual pass revenue?</h3>

In order to determine the effect the increase in price would have on revenue, we have to examine the elasticity of demand of the museum pass.

The point elasticity is 0.62. This means that demand is inelastic. This means that when price increases, there would be little or no change in the quantity demanded. As a result, revenue would increase.

To learn more about price elasticity of demand, please check: brainly.com/question/18850846

You might be interested in

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

The system of equations y = one-fourth x minus 1 and y = negative one-half x minus one-fourth is shown on the graph below. On a

love history [14]

Step-by-step explanation:

<u>Step 1: Convert into expressions</u>

y = one-fourth x minus 1 → $y = \frac{1}{4}x-1$

y = negative one-half x minus one-fourth → $y = -\frac{1}{2}x - \frac{1}{4}$

They intersect at $(1, -\frac{3}{4})$

Answer: Option A, (1, negative three-fourths)

8 0

3 years ago

Read 2 more answers

liberstina [14]

Total is 518.4
Lateral is 174
Hope that helped sorry if it’s wrong

6 0

3 years ago

What is the 24,40,56,72 pattern?

zimovet [89]

This is a arithmetic sequence, so all we need to do is subtract the numbers and find the common difference.

40 - 24 = 16

56 - 40 = 16

72 - 56 = 16

Therefore, the common difference is 16.

Best of Luck!

5 0

3 years ago

Read 2 more answers

Write an equation for the line parallel to the given line that contains C.<br> Help

tangare [24]

There is no photo shown

6 0

3 years ago