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anonymous

  • one year ago

(MEDAL + FAN!) the best highschool is running a fundriser, selling hats with their awesome logo. The hats have been donated by a local merchandise dealer. A survey determines that 50 students are willing to pay $5 for a hat. For every decrease of $.25 in the sale price, 10 more students are expected to purchase a hat. The school must raise at least $390 to meet their goal, but would prefer if as many students as possible purchase a hat. At what price should the hats be sold?

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  1. anonymous
    • one year ago
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    @mathmate @acxbox22

  2. anonymous
    • one year ago
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    heres what i got:

  3. anonymous
    • one year ago
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    let x= # of $.25 decreases in sale price price= #students x price (50+10x)(5-.25x) > 390 -----> the income

  4. anonymous
    • one year ago
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    how do i account for the at least 390 and the most students possible?

  5. anonymous
    • one year ago
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    i think the answer was around $3 if that helps

  6. mathmate
    • one year ago
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    Hint: You have correctly defined x as the \(number\) of 0.25 decreases. You have correctly solved the maximum revenue would be obtained as 7.5 decreases. Unfortunately you have interpreted 7.5 as the dollar amount. Since x can only take on integer values, you need to verify the value of x (around 7.5) that gives the maximum revenue, and the maximum number of students. Check those out, and post if necessary. Btw OS rules stipulate that students should not ask for help from OS for school tests.

  7. anonymous
    • one year ago
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    7.5 is the number of decreases?

  8. mathmate
    • one year ago
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    Please note answer above: "Since x can only take on integer values, you need to verify the value of x (around 7.5) that gives the maximum revenue, and the maximum number of students."

  9. anonymous
    • one year ago
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    x=7 because when i plug it in, give gives 390

  10. anonymous
    • one year ago
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    oh thank, i got $3.25

  11. mathmate
    • one year ago
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    I suggest you read completely and understand answers to your posts. There is still time to reread my previous answer.

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