All real numbers y that are less than 4 or greater than 9

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) Optimal solution: 40.000 units of R and 10.000 of PC = $17.000

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

d) Binding Constaints:

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

Indicate the method you would use to prove the two A's. If no method applies, enter "none".

VikaD [51]

AAS is your answer because you have 30 and 70 degrees next to each other and they are both angels and then 10 which is a side
You would use AAS because that’s the pattern or your triangle

3 0

2 years ago

Read 2 more answers

Simply 15 _ 4 (7d + 9) 2 what is the answer to that

Vilka [71]

Answer:

=2156d+2772

Step-by-step explanation:

3 0

3 years ago

A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their

Sever21 [200]

Answer:

A. Normal

B. Between 40.08 minutes and 43.92 minutes.

C. About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, a normal distribution.

Question B:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{12}{\sqrt{150}} = 1.92$

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.92 = 40.08 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.92 = 43.92 minutes

Between 40.08 minutes and 43.92 minutes.

Question C:

x% confidence interval -> x% will contain the true population mean, (100-x)% wont.

So, 95% confidence interval:

About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

3 0

3 years ago

Using simple linear regression, calculate the trend line for the historical data. say the x axis is april = 1, may = 2, and so o

OverLord2011 [107]

Given a table of historical demand for a product as follows:

$\begin{tabular}
{|c|c|}
 &Demand\\[1ex]
April&60\\
May&55\\
June&75\\July&60\\August&80\\September&75\\
\end{tabular}$

The linear regression equation is given by

$\hat{Y}=bx+a$

where:

$b= \frac{n\Sigma xy-\Sigma x\Sigma y}{n\Sigma x^2-(\Sigma x)^2}$
and
$a= \frac{1}{n} (\Sigma y-b\Sigma x)$

We calculate the required values using the following table, where
April = 1, May = 2, and so on.

$\begin{tabular} 
{|c|c|c|c|}
X &Y&X^2&XY\\[1ex] 
1&60&1&60\\ 
2&55&4&110\\ 
3&75&9&225\\
4&60&16&240\\
5&80&25&400\\
6&75&36&450\\[1ex]
\Sigma X=21&\Sigma Y=405&\Sigma X^2=91&\Sigma XY=1,485 
\end{tabular}$

Thus,

$b= \frac{6(1,485)-21(405)}{6(91)-(21)^2} \\ \\ = \frac{8,910-8,505}{546-441} = \frac{405}{105} \\ \\ \approx3.86$

and

$a= \frac{1}{6} (405-(3.86)(21)) \\ \\ = \frac{1}{6} (405-81)= \frac{1}{6} (324) \\ \\ =54$

Therefore, the trend line for the historical data is given by $\hat{Y}=3.86x+54$

7 0

4 years ago