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3 years ago

A university is trying to determine what price to charge for tickets to football games. At a price of $30 per ticket,

Mathematics

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

The price should be 25 dollars, with this price 50000 people will attend the game.

Step-by-step explanation:

To solve this question we can build a table in which we compare the information taking into account the price of the ticket, the average number of people, the earnings for the sale of the tickets and the earnings of the sales in the concessions and with this determine the maximize revenue that corresponds to the best options for the university so as you can see in the attachment.

To calculate the average people we subtract 10000 per each decrease in $5.

To calculate the earnings per sales we have to multiply the cost of the ticket by the number of average people

To calculate the earnings per concessions we have to multiply the spends in the concessions by the average people

To calculate the total earnings we have to add the earnings per sales with the earnings per concessions.

As you can see, the total earnings for ticket price of $25 and $20 are the same ($150,000) if we increase or decrease the price of the tickets, the total earnings will decrease, then we have to choose the best option.

For maintaining the security and the infrastructure, it will be better to have less people, then the most convenient option will be that in which 50,000 people attend to the game, and for this, the price will be of $25.

Then, the price should be 25 dollars, with this price 50000 people will attend the game.

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2 years ago

Read 2 more answers

I need help finding the area

aniked [119]

<h3>Given :</h3>

Base of triangle = 7 yd
Height of triangle = 10 yd

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Area of triangle

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We know:-

When base and height of triangle is given we use this formula:

$\bigstar \boxed{ \rm Area \: of \: triangle = \frac{base \times height}{2} }$

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So:-

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{base \times height}{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 10}{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times2 }{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times\cancel2 }{\cancel2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times1 }{1} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5$

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$\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2} \\$

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$\therefore \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} { \red{\bf{yd} }^{ \red2} }}$

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$\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$ ]