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Solve the Problem p=15x+200,{x0x500}p = - \frac { 1 } { 5 } x + 200 , \{ x \mid 0 \leq x \leq 500 \}

Question 275

Multiple Choice

Solve the problem.
-The price p and the quantity x sold of a certain product obey the demand equation: p=15x+200,{x0x500}p = - \frac { 1 } { 5 } x + 200 , \{ x \mid 0 \leq x \leq 500 \} What is the revenue to the nearest dollar when 300 units are sold?


A) $110,000
B) $78,000
C) $20,000
D) $42,000

Correct Answer:

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