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Hitman42 [59]
2 years ago
12

Answers to the 2 boxes please :)​

Mathematics
2 answers:
bagirrra123 [75]2 years ago
8 0

Answer:

X = - 1

Y = 0

Step-by-step explanation:

<h3>Hope this helps u... ^_^</h3>
raketka [301]2 years ago
5 0
What the guy on top said
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A concert hall has 8,000 seats and two categories of ticket prices, $29 and $34. Assume that all seats in each category can be s
Greeley [361]

Using a system of equations, it is found that:

  • For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
  • For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
  • For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

--------------------

The variables of the system are:

  • x, which is the number of $29 tickets sold.
  • y, which is the number of $34 tickets sold.

Total of 8,000 seats, all can be sold, thus:

x + y = 8000 \rightarrow x = 8000 - y

--------------------

For a profit of $255,000, we have that:

29x + 34y = 255000

29(8000 - y) + 34y = 255000

5y = 23000

y = \frac{23000}{5}

y = 4600

x = 8000 - 4600 = 3400

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.

--------------------

For a profit of $271,000, we have that:

29x + 34y = 271000

29(8000 - y) + 34y = 271000

5y = 39000

y = \frac{39000}{5}

y = 7800

x = 8000 - 7800= 200

For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.

--------------------

For a profit of $235,000, we have that:

29x + 34y = 235000

29(8000 - y) + 34y = 235000

5y = 3000

y = \frac{3000}{5}

y = 600

x = 8000 - 600 = 7400

For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

A similar problem is given at brainly.com/question/22826010

3 0
2 years ago
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