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chubhunter [2.5K]

2 years ago

10

A furniture company is producing three types of furniture. Product A requires 7 board feet of wood and 4 lbs of wicker. Product

B requires 5 board feet of wood and 5 lbs of wicker. Product C requires 4 board feet of wood and 3 lbs of wicker. There are 3000 board feet of wood available for product and 1400 lbs of wicker. Product A earns a profit margin of $35 a unit, Product B earns a profit margin of $42 a unit, and Product C earns a profit margin of $20 a unit. Formulate the problem as a linear program.

1 answer:

denis23 [38]2 years ago

5 0

Given:

Product A requires 7 board feet of wood and 4 lbs of wicker.

Product B requires 5 board feet of wood and 5 lbs of wicker.

Product C requires 4 board feet of wood and 3 lbs of wicker.

Available wood = 3000 board feet

Available wicker = 1400 lbs

Profit margin of A = $35 per unit

Profit margin of B = $42 per unit

Profit margin of C = $20 per unit

To find:

The linear programming problem for given situation.

Solution:

Let the number of units produced of products A, B and C are x, y and z respectively.

Product A Product B Product C Total

Board feet of wood 7 5 4 3000

wicker 4 5 3 1400

Objective function: Maximize $z=35x+42y+20z$

s.t.,

Board feet of wood : $7x+5y+4x\leq 3000$

Wicker : $4x+5y+3x\leq 1400$

Number of units cannot be negative. So, $x,y,z\geq 0$ .

Therefore, the required LPP is

Maximize $z=35x+42y+20z$

s.t.,

$7x+5y+4x\leq 3000$

$4x+5y+3x\leq 1400$

$x,y,z\geq 0$

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Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

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 $1-\alpha$ .

So it 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. In this problem, $\sigma = 0.25$

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

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

$0.1 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.1\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.1}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}$

$n = 24.01$

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

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

$0.06 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.06\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.06}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}$

$n = 66.7$

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

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

$0.05 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.05\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.05}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}$

$n = 96.04$

Rounding up, 97

8 0

1 year ago

the mean height of the team is 200.3 .a new player joins and raises the mean average height to 201cm. work out the height of the

ivann1987 [24]

Question

The table shows the heights in a basketball team.

Height: 198, 199, 200, 201, 202

a.) Work out the mean height of the team. Round you answer to 1d.p.

That, I have worked out is 200.3

b.) A new player joins the team and raises the mean average height to 201cm. Work out the height of the new player.

Answer:

206 cm

Step-by-step explanation:

The sum of the fiven numbers, 198+199+200+201+202=1000

For a mean of 201, the product should be 201*6=1206

The difference of the two is equivalent to the height of the other person hence 1206-1000=206 cm

Therefore, the height is 206 cm

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1 year ago

In estimating the mean score on a fitness exam, we use an original sample of size n = 30 and a bootstrap distribution containing

ioda

Answer:

C. 67.5 to 72.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The width of the interval is determined by it's margin of error, which is given by the following formula:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

So, as n increases, the margin of error decreases, and the interval gets smaller.

Using 10,000 bootstrap samples for the distribution:

We increase the sample size, which means that the interval gets smaller.

We had 67 to 73, since it got smaller, it will be from a value higher than 67 to a value lower than 73.

So the correct answer is:

C. 67.5 to 72.

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2 years ago

Joan is filling a 2-liter soda bottle with different colors of sand to make an art project for her class. She has 1,000 millilit

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Answer:

0.4 l of purple sand

Step-by-step explanation:

<u>Volume to be filled in:</u>

2 liter = 2000 ml

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1000 ml

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<u>Yellow sand</u>

2.5 deciliters = 2.5*100 ml = 250 ml

<u>Total volume filled:</u>

1000 + 350 + 250 = 1600 ml

<u>Purple sand needed:</u>

2000 - 1600 = 400 ml = 0.4 l

<u>Answer is</u> 0.4 l of purple sand required

6 0

2 years ago

A recreation center holds a soccer game every Saturday morning for older teens. The group agreed that there should be at least f

Lyrx [107]

Answer:

They can play for 1 hour and 20 minutes from the beginning of the game.

Step-by-step explanation:

Start playing with 17 players.

At 1 hour, there are 14 players left.

At 1:05, there are 11 players left.

At 1:10, there are 8 players left.

At 1:15, there are 5 players left.

At 1:20, there are 2 players left, and play stops.

They can play for 1 hour and 20 minutes from the beginning of the game.

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2 years ago