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zhannawk [14.2K]
4 years ago
9

"The table and the graph below each show a different relationship between the same two variables, x and y:

Mathematics
2 answers:
frutty [35]4 years ago
8 0
D 600  because it increases   by the same amount so you just add 55 12 times
sukhopar [10]4 years ago
4 0

Answer:

Option (b) is correct.

The value value of y will be 300 more on the graph than its value in the table when x = 12

Step-by-step explanation:

 Given : a linear graph of given data.

We have to find the value of y when x = 12 .

We find the equation of line for the given graph.

According to table given:

A linear equation is represented as y = mx+c , where m is slope and c  is y- intercept.    

Slope is given as ,

m=\frac{y_2-y_1}{x_2-x_1}

Given : y = 120 when x = 4 that is (4,120)

and  y = 150 when x = 5 that is (5,150)

Substitute, we get,

m=\frac{150-120}{5-4}=30

Thus, slope is 30

Then equation becomes y = 30x + c

and to find c that is y- intercept

Put (5,150) in the above equation , we have,

⇒ 150 = 30× 5 + c

⇒ c = 0

Thus, equation will be y = 30x

To find value of y when x = 12

Put x = 12 in above , we get,

y =  30× 12 = 360      

According to graph

The equation passes to origin  thus, y intercept is 0.

and points are (2, 110)

y= mx

⇒ 110 = 2m

⇒ m = 55

Thus, equation will be y = 55x

Thus, by graph the value  y when x = 12 is

y =  55 × 12 = 660    

Thus, difference is 660 - 360 = 300  

Thus, the value value of y will be 300 more on the graph than its value in the table when x = 12

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Step-by-step explanation:

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Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for reg
Contact [7]

Answer:

a) MAX--> PC (R,P) = 0,3R+ 0,5P

b) <u>Optimal solution</u>: 40.000 units of R and 10.000 of PC = $17.000

c) <u>Slack variables</u>: S3=1000, is the unattended demand of P, the others are 0, that means the restrictions are at the limit.

d) <u>Binding Constaints</u>:

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

Step-by-step explanation:

I will solve it using the graphic method:

First, we have to define the variables:

R : Regular Gasoline

P: Premium Gasoline

We also call:

PC: Profit contributions

A: Grade A crude oil

• R--> PC: $0,3 --> 0,3 A

• P--> PC: $0,5 --> 0,6 A

So the ecuation to maximize is:

MAX--> PC (R,P) = 0,3R+ 0,5P

The restrictions would be:

1. 18.000 A availabe (R=0,3 A ; P 0,6 A)

2. 50.000 capacity

3. Demand of P: No more than 20.000

4. Both P and R 0 or more.

Translated to formulas:

Answer d)

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

To know the optimal solution it is better to graph all the restrictions, once you have the graphic, the theory says that the solution is on one of the vertices.

So we define the vertices: (you can see on the graphic, or calculate them with the intersection of the ecuations)

V:(R;P)

• V1: (0;0)

• V2: (0; 20.000)

• V3: (20.000;20.000)

• V4: (40.000; 10.000)

• V5:(50.000;0)

We check each one in the profit ecuation:

MAX--> PC (R,P) = 0,3R+ 0,5P

• V1: 0

• V2: 10.000

• V3: 16.000

• V4: 17.000

• V5: 15.000

As we can see, the optimal solution is  

V4: 40.000 units of regular and 10.000 of premium.

To have the slack variables you have to check in each restriction how much you have to add (or substract) to get to de exact (=) result.  

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