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3 years ago

Linear Programming. 20) 3x + y ≤ 7; x + 2y ≤ 9; x ≥ 0; y ≥ 0 Maximize the Objective Function: P = 2x + y show the graph

Mathematics

1 answer:

tigry1 [53]3 years ago

8 0

Answer:

The value that maximize the objective function is the point (1,4)

Step-by-step explanation:

we have

$3x+y\leq 7$ ----> inequality A

$x+2y\leq 9$ ----> inequality B

$x\geq 0$ ----> inequality C

$y\geq 0$ ----> inequality D

Using a graphing tool

The solution is the shaded area

see the attached figure

The coordinates of the solution area are

$(0,0),(0,4.5),(1,4),(2.33,0)$

we have

The Objective Function is equal to

$P=2x+y$

To find out the value of x and y that maximize the objective function, substitute each ordered pair of the vertices in the objective function and then compare the results

For (0,0) --------> $P=2(0)+0=0$

For (0,4.5) --------> $P=2(0)+4.5=4.5$

For (1,4) --------> $P=2(1)+4=6$

For (2.33,0) --------> $P=2(2.33)+0=4.66$

The value that maximize the objective function is the point (1,4)

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The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical

Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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.

In this problem, we have that:

$\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538$

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{0.8538}$

$Z = -2.81$

$Z = -2.81$ has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{41 - 38.4}{0.8538}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now $n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08$

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{1.08}$

$Z = -2.22$

$Z = -2.22$ has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

$Z = \frac{X - \mu}{s}$

$Z = \frac{41 - 38.4}{1.08}$

$Z = 2.41$

$Z = 2.41$ has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0

3 years ago

How do I find the ratio of this? It makes no sense to me

Volgvan

Answer:

$\frac{169}{289}$

Step-by-step explanation:

Given 2 similar figures with ratio of sides = a : b, then

ratio of areas = a² : b²

Here ratio of sides = 52 : 68 = 13 : 17 ← in simplest form, thus

ratio of areas = 13² : 17² = 169 : 289 = $\frac{169}{289}$

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3 years ago

How can Sue use the converse of the Pythagorean Theorem to find out if a triangle with side lengths 28, 53, and 45 is a right tr

timama [110]

Answer:

Step-by-step explanation:

28²+45²=53²

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3 years ago

I need help solving this

Tamiku [17]

Answer:

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3 years ago

Read 2 more answers

A carrot has

4vir4ik [10]

Answer:

B- For an equal number of servings, the bananas will have three times as much sugar

Step-by-step explanation:

The complete question is

A carrot has 5 grams of sugar per serving. A banana has 15 grams of sugar per serving. Which statement is true?

A- There will always be more sugar in 10 servings of carrots than in a smaller number of servings of bananas.

B- For an equal number of servings, the bananas will have three times as much sugar.

C- For an equal number of servings, the bananas will always have 10 more grams of sugar.

D- The amount of sugar in two servings of carrots is 30 grams.

Verify each statement

A- There will always be more sugar in 10 servings of carrots than in a smaller number of servings of bananas

The statement is False

Because

In 10 servings of carrots there are 10(5)=50 grams of sugar

In 5 servings of banana there are 5(15)=75 grams of sugar

therefore

50 g < 75 g

B- For an equal number of servings, the bananas will have three times as much sugar

The statement is true

Because

In one serving of carrots there are 5 grams of sugar

In one serving of bananas there are 15 grams of sugar

The ratio of grams of sugar in a serving of bananas to grams of sugar in a serving of carrots is 15:5

Simplify

3:1

therefore

An equal number of servings, the bananas will have three times as much sugar

C- For an equal number of servings, the bananas will always have 10 more grams of sugar.

The statement is False

Because

In 10 servings of carrots there are 10(5)=50 grams of sugar

In 10 servings of banana there are 10(15)=150 grams of sugar

$50+10\neq 150$

D- The amount of sugar in two servings of carrots is 30 grams

The statement is False

Because

In 2 servings of carrots there are 2(5)=10 grams of sugar

$10\neq 30$

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3 years ago