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3 years ago

The constraints of a problem are listed below. What are the vertices of the feasible region?

1 answer:

6 0

The vertices are the following intersection points:

1) y = 0 and x 0 0 => (0,0)

2) x = 0 and 2x - 6y = - 3 => (0, 1/2) = (0, 0.5)

3) 2x + 3y = 6 and 2x - 6y = - 3 => (3/2, 1) = (1.5, 1)

4) y = 0 and 2x + 3y = 6 => (3,0)

Answer: (0,0) , (0, 0.5) (1.5, 1), (3, 0) = the third option.

You might be interested in

Express the distance traveled as a function of the number of hours at the average speed. The morris family is traveling from pro

shusha [124]

Answer:

Dist = 45 m/h * x h

Step-by-step explanation:

We know a speed is a measure of distance over time: speed = dist / time

In this case, we are looking for a distance expressed in terms of time.

So, we only need to modify the formula a bit: dist = speed * time

We don't know the time (which will be a variable), but we know their speed. So, the formula becomes:

Dist = 45 m/h * x h

Enter the number of hours in the formula and that will give you an approximation of the distance traveled within that time (approximation since our result will be rely on an average).

6 0

3 years ago

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

Mamont248 [21]

Answer:

The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$