Real-World Systems of Equations Quick Check
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Question
The senior class at Hartford High School is planning their prom. They’ve narrowed the location down to two choices. The Regal Hotel charges, C, $500 for the ballroom rental fee and $75 per student, s, for the meal. The Greenhill Country Club charges, C, $1,000 for the ballroom rental fee and $65 per student, s, for the meal. Create a pair of linear equations that represent this real-world problem.(1 point)
Responses
Regal Hotel: C=500s+75; Greenhill Country Club: C=1,000s+65
Regal Hotel: upper C equals 500 s plus 75 ; Greenhill Country Club: upper C equals 1,000 s plus 65
Regal Hotel: C=1,000s+75; Greenhill Country Club: C=500s+65
Regal Hotel: upper C equals 1,000 s plus 75 ; Greenhill Country Club: upper C equals 500 s plus 65
Regal Hotel: C=75s+500; Greenhill Country Club: C=65s+1,000
Regal Hotel: upper C equals 75 s plus 500 ; Greenhill Country Club: upper C equals 65 s plus 1,000
Regal Hotel: C=75s+1,000; Greenhill Country Club: C=65s+500
Regal Hotel: C=500s+75; Greenhill Country Club: C=1,000s+65
Regal Hotel: C = 500s + 75; Greenhill Country Club: C = 1,000s + 65
To create a pair of linear equations that represent this real-world problem, we need to assign variables to the different quantities involved. Let's use "C" to represent the total cost and "s" to represent the number of students.
For the Regal Hotel, the equation would be C = 500s + 75. This is because there is a fixed rental fee of $500 for the ballroom and an additional $75 per student for the meal.
For the Greenhill Country Club, the equation would be C = 1000s + 65. Here, the ballroom rental fee is $1000 and there is a meal cost of $65 per student.
So, the correct answer is: Regal Hotel: C=500s+75; Greenhill Country Club: C=1,000s+65.