For the Hawks' 80 basketball games next year, you can buy separate tickets for each game at $9 each, or you can buy a season ticket for $580. At most how many games could you attend at the $9 price before spending more than the cost of a season ticket?

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For the Hawks' 80 basketball games next year, you can buy separate tickets for each game at $9 each, or you can buy a season ticket for $580. At most how many games could you attend at the $9 price before spending more than the cost of a season ticket?

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Answer 1 · 2018-06-26T07:15:33

64 games

Explanation:

$\frac{$ 580}{9} = 64 \frac{4}{9}$ $[$580]/9=64 4/9$
After watching 64 games you would spend less than the season ticket but upon watching the $65^{t h}$ $65^(th)$ game you would have paid more.

Answer 2 · 2018-06-26T07:19:46

65

Explanation:

Let $x$ $x$ be the number of games.

We want to find when $9 x \leq 580$ $9x<=580$

$\frac{9 x}{9} \leq \frac{580}{9}$ $(9x)/9<=580/9$

$x \leq 64 + \frac{4}{9}$ $x<=64+4/9$

$:.$ The 65th game is the first one more expensive to buy individually.

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For the Hawks' 80 basketball games next year, you can buy separate tickets for each game at $9 each, or you can buy a season ticket for $580. At most how many games could you attend at the $9 price before spending more than the cost of a season ticket?

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