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

Question

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

Answer 1

To create a pair of linear equations representing the costs of the two venues for the prom, we can summarize the cost components for each venue.

For the Regal Hotel:

• The ballroom rental fee is $500.
• The cost per student for the meal is $75.

The equation for the Regal Hotel can be expressed as: \[ C = 75s + 500 \]

For the Greenhill Country Club:

• The ballroom rental fee is $1,000.
• The cost per student for the meal is $65.

The equation for the Greenhill Country Club can be expressed as: \[ C = 65s + 1000 \]

Thus, the correct set of equations is:

• Regal Hotel: \( C = 75s + 500 \)
• Greenhill Country Club: \( C = 65s + 1000 \)

So, the correct response is:

Regal Hotel: C = 75s + 500; Greenhill Country Club: C = 65s + 1000.

Answer 2

The Jackson Jammers are giving away bags and caps at their next game. The bags, b
, cost them $10 each, and the caps, c
, cost them $5 each. They plan to give away 2,000 total items and have a budget of $12,500. Write the pair of linear equations that model this situation.(1 point)
Responses

b+c=2,000
; 5b+10c=12,500
b plus c equals 2,000 ; 10 b plus 10 c equals 12,500

b+c=12,500
; 5b+10c=2,000
b plus c equals 12,500 ; 5 b plus 10 c equals 2,000

b+c=2,000
; 10b+5c=12,500
b plus c equals 2,000 ; 10 b plus 5 c equals 12,500

b+c=12,500
; 10b+5c=2,000

Answer 3

To model the situation with the bags and caps, we can set up two linear equations based on the information provided.

1. Total Items Equation: The total number of bags \( b \) and caps \( c \) they plan to give away amounts to 2,000 items. \[ b + c = 2000 \]

2. Budget Equation: The cost of the bags is $10 each, and the cost of the caps is $5 each. Their budget is $12,500. Thus, the total cost can be represented as: \[ 10b + 5c = 12500 \]

Therefore, the pair of linear equations that model this situation is:

• \( b + c = 2000 \)
• \( 10b + 5c = 12500 \)

The correct response is: b + c = 2,000; 10b + 5c = 12,500.

Answer 4

Solve this real-world problem using a system of equations. The O’Donnells are having a birthday party for their son, and they plan to have a clown bring balloons to the party. Balloon Express charges $2 per balloon and $10 for delivery. Balloon Mania charges $1.50 per balloon and $20 for delivery. If the O’Donnells plan to order 30 balloons, which company should they choose, and how much will they save?(1 point)
Responses

Balloon Express; $25
Balloon Express; $25

Balloon Express; $5
Balloon Express; $5

Balloon Mania; $5
Balloon Mania; $5

Balloon Mania; $25 

Answer 5

To determine which company the O'Donnells should choose for the balloons and how much they will save, we can set up equations for the total cost from each company based on their pricing structures and the number of balloons they plan to order (30).

Costs Calculation:

1. Balloon Express:

   • Cost per balloon: $2
   • Delivery fee: $10
   • Total cost, \( C_E \), for 30 balloons: \[ C_E = (2 \times 30) + 10 = 60 + 10 = 70 \]

2. Balloon Mania:

   • Cost per balloon: $1.50
   • Delivery fee: $20
   • Total cost, \( C_M \), for 30 balloons: \[ C_M = (1.50 \times 30) + 20 = 45 + 20 = 65 \]

Comparison:

• Total cost from Balloon Express: $70
• Total cost from Balloon Mania: $65

Savings Calculation:

To find out how much the O'Donnells will save by choosing Balloon Mania over Balloon Express: \[ \text{Savings} = C_E - C_M = 70 - 65 = 5 \]

Conclusion:

The O'Donnells should choose Balloon Mania, and they will save $5.

The correct response is: Balloon Mania; $5.

Answer 6

The Spanish Club at Santa Fe Middle School wants to sell mini quiches to raise money for a field trip. The culinary department at their school will make the quiches for $30 plus $0.50 per quiche for the ingredients. The Spanish Club wants to sell each quiche for $2. How many quiches will they need to sell in order to break even?(1 point)
Responses

45 quiches
45 quiches

12 quiches
12 quiches

20 quiches
20 quiches

75 quiches

Answer 7

To determine how many quiches the Spanish Club needs to sell to break even, we first need to set up the equations representing the costs and the revenue.

Costs:

The total cost \( C \) of making \( q \) quiches includes a fixed cost of $30 and a variable cost of $0.50 per quiche. This can be expressed as: \[ C = 30 + 0.5q \]

Revenue:

The revenue \( R \) from selling \( q \) quiches, each for $2, can be expressed as: \[ R = 2q \]

Break-Even Point:

To find the break-even point, we set the total cost equal to the total revenue: \[ 30 + 0.5q = 2q \]

Now, we can solve for \( q \):

1. Subtract \( 0.5q \) from both sides: \[ 30 = 2q - 0.5q \] \[ 30 = 1.5q \]

2. Divide both sides by 1.5: \[ q = \frac{30}{1.5} = 20 \]

Conclusion:

The Spanish Club will need to sell 20 quiches to break even.

The correct response is: 20 quiches.

Answer 8

The Baynard High School Boosters are selling hot dogs and pretzels at the concession stand. They have 300 hot dogs and pretzels in stock and want to earn an income of $800. They plan to sell the hot dogs for $4 each and the pretzels for $2 each. How many hot dogs and how many pretzels do the boosters need to sell to reach their goal of $800?(1 point)
Responses

200 hot dogs and 100 pretzels
200 hot dogs and 100 pretzels

100 hot dogs and 200 pretzels
100 hot dogs and 200 pretzels

50 hot dogs and 250 pretzels
50 hot dogs and 250 pretzels

250 hot dogs and 50 pretzels
250 hot dogs and 50 pretzels

Answer 9

To determine how many hot dogs and pretzels the Baynard High School Boosters need to sell in order to reach their income goal of $800 while having a total stock of 300 items, we can set up a system of equations.

Let:

• \( h \) = number of hot dogs
• \( p \) = number of pretzels

Given Information:

1. Total items equation:
   The total number of hot dogs and pretzels is 300. \[ h + p = 300 \tag{1} \]

2. Income equation:
   The income from selling hot dogs at $4 each and pretzels at $2 each must equal $800. \[ 4h + 2p = 800 \tag{2} \]

Solving the System of Equations:

Let's start with equation (1): \[ p = 300 - h \tag{3} \]

Now, substitute equation (3) into equation (2): \[ 4h + 2(300 - h) = 800 \]

Expanding this gives: \[ 4h + 600 - 2h = 800 \]

Combine like terms: \[ 2h + 600 = 800 \]

Now, isolate \( h \): \[ 2h = 800 - 600 \] \[ 2h = 200 \] \[ h = 100 \]

Using equation (3) to find \( p \): \[ p = 300 - h = 300 - 100 = 200 \]

Conclusion:

To reach their goal of $800, the Boosters need to sell 100 hot dogs and 200 pretzels.

The correct response is: 100 hot dogs and 200 pretzels.

Answer 10

Julie is asked to solve the system below by elimination. Which of the following steps would be the best way to begin?

i. 2x+y=−3
ii. 3x−2y=2
(1 point)
Responses

Multiply equation i by 2 and multiply equation ii by 3
Multiply equation i by 2 and multiply equation ii by 3

Multiply equation ii by 2
Multiply equation ii by 2

Multiply equation i by 2
Multiply equation i by 2

Multiply equation ii by 2 and multiply equation i by -2
Multiply equation ii by 2 and multiply equation i by -2

Answer 11

To solve the given system of equations using the elimination method, you want to eliminate one of the variables by ensuring that the coefficients of that variable are opposites when adding the equations together.

The system is:

1. \( 2x + y = -3 \) (equation i)
2. \( 3x - 2y = 2 \) (equation ii)

Options Consideration:

1. Multiply equation i by 2 and multiply equation ii by 3:

   • This could make the coefficients of \( y \) equal (or make them opposites) but is not the most efficient starting point.

2. Multiply equation ii by 2:

   • This changes equation ii to \( 6x - 4y = 4 \). This does not directly help eliminate \( y \) since it doesn't match the coefficient of \( y \) from equation i.

3. Multiply equation i by 2:

   • This would yield \( 4x + 2y = -6 \), which is not useful as it doesn't aid in eliminating \( y \) effectively.

4. Multiply equation ii by 2 and multiply equation i by -2:

   • If we multiply equation ii by 2: \( 6x - 4y = 4 \)
   • If we multiply equation i by -2: \( -4x - 2y = 6 \)
   • Adding these two results would help eliminate \( y \): \[ 6x - 4y + (-4x - 2y) = 4 + 6 \] which simplifies to: \[ 2x - 6y = 10 \]

Conclusion:

The best way to begin solving this system using elimination is: Multiply equation ii by 2 and multiply equation i by -2.

So, the correct response is: Multiply equation ii by 2 and multiply equation i by -2.