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stealth61 [152]
3 years ago
7

The high school jazz band is selling homemade leather bracelets at a local craft fair to raise money for a trip. The group has a

$200 budget to spend on supplies, which is enough to make 260 bracelets. The group is charging $2 per bracelet at the craft fair.
Part A Suppose b is the number of bracelets sold, and f(b) is the profit the band makes on the sales. What are the independent and dependent variables in this situation?
Part B Write a function modeling the relationship.
Part C Suppose the group sells exactly 100 bracelets. What will be the profit from the sales?
Part D Find the following for this situation: the minimum and maximum values in the domain the maximum value in the range
Mathematics
2 answers:
amid [387]3 years ago
4 0
<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ A)  The number of bracelets sold is the independent variable and the profit is the dependent variable

B) f(b) = 2b - 200

C) Substitute the value in:

f(b) = 2(100) - 200

The profit would be $0

D) i don't actually know this part sorry :(

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

raketka [301]3 years ago
4 0

Answer:

Which statements about this situation are true?

Select all the correct answers.

THE  CORRECT ANSWER ARE B AND D

The maximum value in the range is $200.

The maximum value in the range is $320.

The maximum value in the domain is 200.

The minimum value in the domain is 0.

Step-by-step explanation:

The domain is defined by b, the number of bracelets sold.

The minimum value in the domain is 0, which represents no bracelets sold.

The maximum value in the domain is 260, which represents the largest number of bracelets the group can make, and the largest number they could sell.

The range is defined by f(b), the amount of profit on the bracelets.

To find the maximum value in the range, we find f(260), the profit on selling the maximum in the domain.

Substitute 260 for b in f(b) = 2b – 200:

f(260) = 2(260) – 200

f(260) = 320

The maximum value in the range is $320.

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