Solve the problem.
-An airline charter service charges a fare per person of $350 plus $20 for each unsold seat. The airplane holds 125 passengers. Let x represent the number of unsold seats and write an expression for the total revenue R for a
Charter flight.
A) $R ( x ) = ( 125 - x ) ( 350 + 20 x )$ or $43,750 + 2,150 x - 20 x ^ { 2 }$
B) $R ( x ) = 125 ( 350 + 20 x )$ or $43,750 + 2,500 x$
C) $R ( x ) = x ( 350 + 20 x )$ or $350 x + 20 x ^ { 2 }$
D) $R ( x ) = ( 125 - x ) ( 350 + 20 x )$ or $43,750 + 2,500 x - 20 x ^ { 2 }$

Question

Solve the problem.
-An airline charter service charges a fare per person of $350 plus $20 for each unsold seat. The airplane holds 125 passengers. Let x represent the number of unsold seats and write an expression for the total revenue R for a
Charter flight. 
A) $R ( x ) = ( 125 - x ) ( 350 + 20 x )$ or $43,750 + 2,150 x - 20 x ^ { 2 }$
B) $R ( x ) = 125 ( 350 + 20 x )$ or $43,750 + 2,500 x$
C) $R ( x ) = x ( 350 + 20 x )$ or $350 x + 20 x ^ { 2 }$
D) $R ( x ) = ( 125 - x ) ( 350 + 20 x )$ or $43,750 + 2,500 x - 20 x ^ { 2 }$

Quizplus · Accepted Answer

The total revenue R for a charter flight is calculated by multiplying the fare per person by the number of passengers (125 - x) and adding $20 for each unsold seat (x). The correct expression is $R(x) = (125 - x)(350 + 20x)$, which simplifies to $43,750 + 2,150x - 20x^2$.

Solve the Problem R(x)=(125−x)(350+20x)R ( x ) = ( 125 - x ) ( 350 + 20 x )R(x)=(125−x)(350+20x)

Solve the Problem $R ( x ) = ( 125 - x ) ( 350 + 20 x )$