Question

traveling carnivals move from town to town, staying for a limited number of days before moving to the next stop. the management of a certain carnival knows that, each time it opens in a new town, it can expect to bring in about $15,000 in revenue the first night. each night after the first, revenue will be about 75% of the previous nights revenue. to the nearest dollar, about how much revenue can the carnival anticipate on its 5th night in town? on the 5th night, the carnival can expect $a in revenue.

Explanation:

Step1: Identify the formula

This is a geometric - sequence problem. The formula for the $n$th term of a geometric sequence is $a_n=a_1r^{n - 1}$, where $a_1$ is the first - term, $r$ is the common ratio, and $n$ is the term number.

Step2: Determine the values of $a_1$, $r$, and $n$

We are given that $a_1 = 15000$ (revenue on the first night), $r=0.75$ (since each night's revenue is 75% or 0.75 of the previous night's revenue), and $n = 5$ (we want to find the revenue on the 5th night).

Step3: Substitute the values into the formula

$a_5=15000\times(0.75)^{5 - 1}=15000\times(0.75)^4$.

Step4: Calculate $(0.75)^4$

$(0.75)^4=0.75\times0.75\times0.75\times0.75 = 0.31640625$.

Step5: Calculate $a_5$

$a_5=15000\times0.31640625 = 4746.09375\approx4746$.

Answer:

$4746$