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Tanya [424]
3 years ago
7

1,2,3 or 4? help please and no I can't just guess till I get it right please help me

Mathematics
1 answer:
ivolga24 [154]3 years ago
5 0
The last option would be correct! This is because she reads at least 50 hours a month plus x (the number of days in the month, which varies) so y would be greater than that. But, she could read exactly 50 hours plus exactly the number of days in the month so equal to is also an option here. :)
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3*4*4=48
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CAN SOMEONE HELP DO #1 PLEASE
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Answer:

  B.  $450

Step-by-step explanation:

The amount of commission is the commission rate (3%) times the amount of sales to which it is applied ($5000). Garrett's commission last week would be ...

  0.03 × $5000 = $150

He is also paid $300 each week in addition to his commission, so his total pay last week was ...

  pay = salary + commission

  pay = $300 +150 = $450

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Mariulka [41]

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2 years ago
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.  

3 0
3 years ago
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