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igor_vitrenko [27]

3 years ago

6

The total number of people at a football game was 5600. Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

If the total amount of money received for the tickets was $134000, how many of each kind of ticket were sold

1 answer:

rewona [7]3 years ago

3 0

Answer:

1100 field-side tickets and 4500 end-zone tickets.

Step-by-step explanation:

Let x represent number of field side tickets and y represent number of end-zone tickets.

We have been given that the total number of people at a football game was 5600. We can represent this information in an equation as:

$x+y=5600...(1)$

$y=5600-x...(1)$

We are also told that Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

Cost of x field side tickets would be $40x$ and cost of y end-zone tickets would be $20y$ .

The total amount of money received for the tickets was $134000. We can represent this information in an equation as:

$40x+20y=134000...(2)$

Upon substituting equation (1) in equation (2), we will get:

$40x+20(5600-x)=134000$

$40x+112000-20x=134000$

$20x+112000=134000$

$20x+112000-112000=134000-112000$

$20x=22000$

$\frac{20x}{20}=\frac{22000}{20}$

$x=1100$

Therefore, 1100 field side tickets were sold.

Upon substituting $x=1100$ in equation (1), we will get:

$y=5600-1100$

$y=4500$

Therefore, 4500 end-zone tickets were sold.

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Part iii

We want this probability:

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And replacing we got:

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