Traveling carnivals usually shut down for the winter months. Suppose a traveling carnival decided to instead remain open for business indefinitely in its last stop of the season. Assume its usual revenue projections hold($15,000 on the first nigh, and each night's revenue after the first night will be about 75% of the previous night revenue). To the nearest dollar, what is the maximum amount of revenue the carnival can expect in the town?

A geometric progression is defined as a sequence in which any element after the first is obtained by multiplying the previous element by a constant which is called a common ratio denoted by r.

For example, the sequence 1, 4, 16, 64,… is a geometric sequence with a common ratio of r = 4.

We are told that Revenue in the first night = $15000

Now revenue of new night is 75% of the previous night. Thus;

2nd night = 15000 × 75/100 = $11250

3rd night = $11250 × 75/100 = 8437.5

Therefore, the given geometric sequence is ;

15000, 11250, 8437.5 ...

First term; a = $15000

Common ratio (r) = 11250/15000 = 0.75

The formula for the nth term of a GP is given by ;

aₙ = arⁿ ⁻ ¹

Thus, revenue for the 5th night is;

a₅ = ar⁵⁻¹

a₅ = 15000(0.75)⁴

a₅ = 4746.093

We conclude that the revenue in the 5th night in two will be $4746.093.

Read more about Geometric Progression at; https://brainly.com/question/24643676

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