Question:
Mancini's Pizzeria sells four types of pizza crust. Last week, the owner tracked the number sold of each type, and this is what he found.
Type of Crust Number Sold
Thin crust 312
Thick crust 245
Stuffed crust 179
Pan style 304
Based on this information, of the next 4500 pizzas he sells, how many should he expect to be thick crust? Round your answer to the nearest whole number. Do not round any intermediate calculations.
Answer:
1060 thick crusts
Step-by-step explanation:
Given
The above table
Required
Expected number of thick crust for the next 4500
For last week data, calculate the proportion of thick crust sold
For the next 4500;
The expected number of thick crust is (E(x)):
System A:
6x + y = 2
-x - y = -3
System B:
2x - 3y = -10
-x-y = -3
Solve:
System A:
6x + y = 2
y = 2 - 6x
-x - (2-6x) = -3
-x - 2 + 6x = -3
5x = -3 + 2
5x = -1
x = -1/5
y = 2 - 6(-1/5)
y = 2 + 6/5
y = 2 + 1.2
y = 3.2 System A: x = -1/5 or -0.2 ; y = 3 1/5 or 3.2
System B:
2x - 3y = -10
2x = -10 + 3y
x = -5 + 1.5y
-x - y = -3
-(-5 + 1.5y) -y = -3
5 - 1.5y - y = -3
-2.5y = -3 - 5
-2.5y = -8
y = 3.2
x = -5 + 1.5(3.2)
x = -5 + 4.8
x = -0.2 System B: x = -0.2 ; y = 3.2
<span>B) They will have the same solution because the first equations of both the systems have the same graph.</span>
Answer:
15 ounces per box
Step-by-step explanation:
The rate StartFraction 165 ounces Over 11 boxes EndFraction describes the relationship between the number of boxes and the weight of the crackers in the boxes. What is the weight, in ounces, of one box?
Total weight of crackers in the boxes = 165 ounces
Total number of boxes = 11 boxes
What is the weight, in ounces, of one box?
Weight per box of crackers =
Total weight of crackers in the boxes / Total number of boxes
= 165 ounces / 11 boxes
= 15 ounces per box
The weight, in ounces, of one box is 15 ounces
Answer:
I say its number 2
Step-by-step explanation:
The table is attached as a figure
The given equation is ⇒⇒⇒⇒ y = 2x
To solve this equation, we need to pick numbers from the table then this number will be substituted into the equation to find y
we need 3 solutions . so, we need to pick 3 numbers of x
From the table, let us choose x = 0
y = 2x
y = 2 * (0)
y = 0
From the table, let choose second value of x such as x = 1
y = 2 * (1)
y = 2
From the table, let choose third value of x such as x = -1
y = 2 * (-1)
y = -2
So, the picked three solutions are y = 0 at x = 0y = -2 at x = -1y = 2 at x = 1
