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Gelneren [198K]
3 years ago
10

Tickets to a football game cost 5$ for students and 7$ for adults. write and solve a system of equations to determine the number

to student tickets sold and the number of adult tickets sold at the most recent home football game if a total of 680 people purchased tickets and the ticket booth sold 3,914 worth of tickets. Does anyone know the equation? Tysm in advance tho if you help!
Mathematics
1 answer:
stepan [7]3 years ago
8 0

Answer:

x + y = 680

5x + 7y = 3914

Step-by-step explanation:

Let the amount of students who purchased tickets equal x.

Let the amount of adults who purchased tickets equal y.

If you add the amount of adults and students, you get the total amount of people:

x + y = 680

Also, to find the amount students spent on tickets, you multiply the ticket price by the amount of students to get 5x. And to find the amount adults spent on tickets, you multiply the ticket price by the amount of adults to get 7y. By adding the two amount, you get the total amount of money:

5x + 7y = 3914

These are the two equations:

x + y = 680

5x + 7y = 3914

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a) 240 ways

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a) The number of ways one component of each type can be selected =

\left(\begin{array}{ccc}5\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right)

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b) If both the receiver and compact disk are to be sony.

In the receiver, the purchaser was offered 1 Sony, also in the CD(compact disk) player the purchaser was offered 1 Sony.

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\left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right)

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c) If none is to be Sony.

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CD player has 1 Sony = 4 - 1 = 3

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Turntable has 1 sony = 4 - 1 = 3

Therefore, the number of ways can be selected if none is to be sony:

\left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right)

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ii) Probability of selecting exactly one sony component =

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= \frac{(1*3*3*3)+(4*1*3*3)+(4*3*3*1)}{240}

\frac{27 + 36 + 36}{240} = \frac{99}{240} = 0.4125

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