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)

Question

Responses

Regal Hotel: C=75s+1,000; Greenhill Country Club: C=65s+500
Regal Hotel: upper C equals 75 s plus 1,000 ; Greenhill Country Club: upper C equals 65 s plus 500

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=500s+75; Greenhill Country Club: C=1,000s+65

Answer 1

To create linear equations based on the information given about the costs at the Regal Hotel and the Greenhill Country Club, we can use the following definitions:

1. Regal Hotel:

   • Ballroom rental fee: $500
   • Meal cost per student: $75
   • Total cost (C) can be represented as: \[ C = 75s + 500 \]

2. Greenhill Country Club:

   • Ballroom rental fee: $1,000
   • Meal cost per student: $65
   • Total cost (C) can be represented as: \[ C = 65s + 1,000 \]

Therefore, the correct pair of linear equations is:

• Regal Hotel: \( C = 75s + 500 \)
• Greenhill Country Club: \( C = 65s + 1,000 \)

So the response is:

Regal Hotel: \( C = 75s + 500 \); Greenhill Country Club: \( C = 65s + 1,000 \)