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 20 quiches 20 quiches 12 quiches 12 quiches 75 quiches 75 quiches 45 quiches
To break even, the Spanish Club needs to cover the cost of making the quiches, which is $30 + $0.50 per quiche. Let's call the number of quiches they need to sell x.
The total cost of making the quiches is $30 + ($0.50 * x).
The Spanish Club wants to sell each quiche for $2, so the total revenue from selling x quiches is $2 * x.
To break even, the revenue from selling the quiches should equal the cost of making them:
$2 * x = $30 + ($0.50 * x)
Now we can solve for x:
$2 * x - $0.50 * x = $30
1.5 * x = $30
x = $30 / 1.5
x = 20
Therefore, the Spanish Club will need to sell 20 quiches in order to break even.
To answer this question, we need to calculate the total cost of making the quiches and then determine how many quiches need to be sold at the selling price of $2 each to cover that cost.
First, let's calculate the cost per quiche. The culinary department charges $30 for making the quiches plus an additional $0.50 per quiche for the ingredients. Therefore, the cost per quiche is $30 + $0.50 = $30.50.
Now, we need to determine the number of quiches that need to be sold to cover this cost. To do this, we divide the total cost by the selling price per quiche. The total cost is $30.50, and the selling price is $2. So, the number of quiches needed to break even is $30.50 ÷ $2 = 15.25.
Since it's not possible to sell a fraction of a quiche, we need to round up the number of quiches to the nearest whole number. Therefore, the Spanish Club will need to sell at least 16 quiches to break even.
To determine the number of quiches the Spanish Club needs to sell to break even, we need to calculate the total cost of making the quiches and then divide it by the selling price of each quiche.
The cost to make the quiches is $30 plus $0.50 per quiche for the ingredients. Let's say the number of quiches they need to sell to break even is x.
Total cost = $30 + $0.50x
Selling price per quiche = $2
To break even, the total cost should equal the total revenue:
$30 + $0.50x = $2x
Now, we can solve for x:
$30 = $1.50x
x = $30 / $1.50
x = 20
Therefore, the Spanish Club will need to sell 20 quiches in order to break even.
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 100 hot dogs and 200 pretzels 100 hot dogs and 200 pretzels 250 hot dogs and 50 pretzels 250 hot dogs and 50 pretzels 50 hot dogs and 250 pretzels 50 hot dogs and 250 pretzels 200 hot dogs and 100 pretzels 200 hot dogs and 100 pretzels
Let's say the number of hot dogs the Boosters need to sell is x, and the number of pretzels they need to sell is y.
The selling price for each hot dog is $4, so the revenue from selling hot dogs is 4x.
Similarly, the selling price for each pretzel is $2, so the revenue from selling pretzels is 2y.
To reach their goal of $800, the total revenue from hot dogs and pretzels should equal $800:
4x + 2y = 800
We also know that the total number of hot dogs and pretzels they have in stock is 300:
x + y = 300
Now we have a system of equations. We can solve it to find the values of x and y.
From the second equation, we can express x in terms of y as: x = 300 - y.
Substituting this value of x into the first equation, we get:
4(300 - y) + 2y = 800
1200 - 4y + 2y = 800
-2y = -400
y = 200
Substituting this value of y back into x = 300 - y, we get:
x = 300 - 200
x = 100
Therefore, the Boosters need to sell 100 hot dogs and 200 pretzels to reach their goal of $800.
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 100 hot dogs and 200 pretzels 100 hot dogs and 200 pretzels 250 hot dogs and 50 pretzels 250 hot dogs and 50 pretzels 50 hot dogs and 250 pretzels 50 hot dogs and 250 pretzels 200 hot dogs and 100 pretzels 200 hot dogs and 100 pretzels
To calculate the number of hot dogs and pretzels the Boosters need to sell to reach their goal of $800, we can set up a system of equations and solve it.
Let's say the number of hot dogs they need to sell is H, and the number of pretzels they need to sell is P.
The selling price for each hot dog is $4, so the revenue from selling hot dogs is 4H.
Similarly, the selling price for each pretzel is $2, so the revenue from selling pretzels is 2P.
We know that the total number of hot dogs and pretzels they have in stock is 300:
H + P = 300
We also know that the total revenue they want to earn is $800:
4H + 2P = 800
Now we can solve this system of equations to find the values of H and P.
Multiplying the first equation by 2, we get:
2H + 2P = 600
Subtracting this equation from the second equation, we get:
4H + 2P - (2H + 2P) = 800 - 600
2H = 200
H = 100
Substituting this value of H back into the first equation, we get:
100 + P = 300
P = 200
Therefore, the Boosters need to sell 100 hot dogs and 200 pretzels to reach their goal of $800.
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
To begin solving the system by elimination, the best way would be to multiply equation i by 2 and multiply equation ii by 3.
This will allow us to create opposite coefficients for the y term in both equations, so that when we add the two equations together, the y terms will cancel out. This is a common technique in the elimination method.