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Let's analyze each of the given statements based on the quadratic equation [tex]\( y = -6x^2 + 100x - 180 \)[/tex], where [tex]\( x \)[/tex] represents the selling price of each soccer ball and [tex]\( y \)[/tex] represents the daily profit.
1. If the store charges [tex]$\$[/tex] 5[tex]$ per soccer ball, it can expect to sell 170 of them. - To verify this, we substitute \( x = 5 \) into the quadratic equation to calculate the profit: \[ y = -6(5)^2 + 100(5) - 180 \] \[ y = -6(25) + 500 - 180 \] \[ y = -150 + 500 - 180 \] \[ y = 170 \] Thus, the profit when the store charges \$[/tex]5 per soccer ball is indeed \[tex]$170. 2. If the store charges $[/tex]\[tex]$ 8$[/tex] per soccer ball, it makes a daily profit of [tex]$\$[/tex] 236[tex]$. - To verify this, we substitute \( x = 8 \) into the quadratic equation to calculate the profit: \[ y = -6(8)^2 + 100(8) - 180 \] \[ y = -6(64) + 800 - 180 \] \[ y = -384 + 800 - 180 \] \[ y = 236 \] Thus, the profit when the store charges \$[/tex]8 per soccer ball is \[tex]$236. 3. If the store sells 12 soccer balls, it makes \$[/tex]156 in profit.
- This statement involves the number of soccer balls sold, but our original quadratic equation relates to the selling price, not the number of soccer balls sold. Assuming a misinterpretation:
[tex]\[ y = profit(12) \][/tex]
Substituting [tex]\( x = 12 \)[/tex] into the equation:
[tex]\[ y = -6(12)^2 + 100(12) - 180 \][/tex]
[tex]\[ y = -6(144) + 1200 - 180 \][/tex]
[tex]\[ y = -864 + 1200 - 180 \][/tex]
[tex]\[ y = 156 \][/tex]
Thus, if the store's equation is assumed based on pricing and not actual balls sold, it makes \[tex]$156 when considering evaluation for \( x=12 \). 4. The greater the selling price, the greater the profit. - To verify this, we consider the nature of the quadratic equation \( y = -6x^2 + 100x - 180 \): The quadratic term \(-6x^2\) indicates that the equation represents a parabola opening downwards. This means there is a maximum profit at a certain price, but not an ever-increasing profit: Substituting various reasonable prices will show the maximum and decreasing trend. Additionally, the Python output confirms this pattern is not increasing consistently: \[ greater\_profit\_check = False \] In summary, the true statements based on our calculations are: - If the store charges \$[/tex]5 per soccer ball, it can expect to make a profit of \[tex]$170. - If the store charges \$[/tex]8 per soccer ball, it makes a daily profit of \[tex]$236. - If evaluated at selling price $[/tex]12, the resulting profit calculation indicates \$156.
- The statement that greater prices always yield greater profits is false.
1. If the store charges [tex]$\$[/tex] 5[tex]$ per soccer ball, it can expect to sell 170 of them. - To verify this, we substitute \( x = 5 \) into the quadratic equation to calculate the profit: \[ y = -6(5)^2 + 100(5) - 180 \] \[ y = -6(25) + 500 - 180 \] \[ y = -150 + 500 - 180 \] \[ y = 170 \] Thus, the profit when the store charges \$[/tex]5 per soccer ball is indeed \[tex]$170. 2. If the store charges $[/tex]\[tex]$ 8$[/tex] per soccer ball, it makes a daily profit of [tex]$\$[/tex] 236[tex]$. - To verify this, we substitute \( x = 8 \) into the quadratic equation to calculate the profit: \[ y = -6(8)^2 + 100(8) - 180 \] \[ y = -6(64) + 800 - 180 \] \[ y = -384 + 800 - 180 \] \[ y = 236 \] Thus, the profit when the store charges \$[/tex]8 per soccer ball is \[tex]$236. 3. If the store sells 12 soccer balls, it makes \$[/tex]156 in profit.
- This statement involves the number of soccer balls sold, but our original quadratic equation relates to the selling price, not the number of soccer balls sold. Assuming a misinterpretation:
[tex]\[ y = profit(12) \][/tex]
Substituting [tex]\( x = 12 \)[/tex] into the equation:
[tex]\[ y = -6(12)^2 + 100(12) - 180 \][/tex]
[tex]\[ y = -6(144) + 1200 - 180 \][/tex]
[tex]\[ y = -864 + 1200 - 180 \][/tex]
[tex]\[ y = 156 \][/tex]
Thus, if the store's equation is assumed based on pricing and not actual balls sold, it makes \[tex]$156 when considering evaluation for \( x=12 \). 4. The greater the selling price, the greater the profit. - To verify this, we consider the nature of the quadratic equation \( y = -6x^2 + 100x - 180 \): The quadratic term \(-6x^2\) indicates that the equation represents a parabola opening downwards. This means there is a maximum profit at a certain price, but not an ever-increasing profit: Substituting various reasonable prices will show the maximum and decreasing trend. Additionally, the Python output confirms this pattern is not increasing consistently: \[ greater\_profit\_check = False \] In summary, the true statements based on our calculations are: - If the store charges \$[/tex]5 per soccer ball, it can expect to make a profit of \[tex]$170. - If the store charges \$[/tex]8 per soccer ball, it makes a daily profit of \[tex]$236. - If evaluated at selling price $[/tex]12, the resulting profit calculation indicates \$156.
- The statement that greater prices always yield greater profits is false.
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