Sheila can ride her bicycle 6000 meters in 15 minutes. How far can she ride her bicycle in 2 minutes.

Answers

Answer 1
Answer:

Answer: about 500

Step-by-step explanation:

Answer 2
Answer:

Answer:

Step-by-step explanation:

800 meters


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Sanderson Manufacturing produces ornate, decorative wood frame doors and windows. Each item produced goes through three manufacturing processes: cutting, sanding, and finishing. Each door produced requires 1 hour in cutting, 30 minutes in sanding, and 30 minutes in finishing. Each window requires 30 minutes in cutting, 45 minutes in sanding, and 1 hour in finishing. In the coming week Sanderson has 40 hours of cutting capacity available, 40 hours of sanding capacity, and 60 hours of finishing capacity. Assume all doors produced can be sold for a profit of $500 and all windows can be sold for a profit of $400.Required:
a. Formulate an LP model for this problem.
b. Sketch the feasible region.
c. What is the optimal solution?

Answers

Answer:

Let X1 be the number of decorative wood frame doors and X2 be the number of windows.  

The profit earned from selling each door is $500 and the profit earned from selling of each window is $400.  

The Sanderson Manufacturer wants to maximize their profit. So for this model, the objective function is

Max: 500X1 + 400X2

Now the total time available for cutting of door and window are 2400 minutes.  

so the time taken in cutting should be less than or equal to 2400.  

60X1 + 30X2 ≤ 2400  

The total available time for sanding of door and window are 2400 minutes. Therefore, the time taken in sanding will be less than or equal to 2400.   30X1 + 45X2 ≤ 2400  

The total time available for finishing of door and window is 3600 hours. Therefore, the time taken in finishing will be less than or equal to 3600. 30X1 + 60X2 ≤ 3600  

As the number of decorative wood frame door and the number of windows cannot be negative.  

Therefore, X1, X2 ≥ 0

so the questions

a)

The LP mode for this model is;

Max: 500X1 + 400X2  

Subject to:  

60X1 + 30X2 ≤ 2400  

]30X1 +45X2 ≤ 2400  

30X1 + 60X2 ≤ 3600  

X1, X2 ≥ 0  

b) Plot the graph of the LP  

Max: 500X1+ 400X2  

Subject to:  

60X1 + 30X2 ≤ 2400  

30X1 + 45X2 ≤ 2400  

30X1 + 60X2 ≤ 3600

X1,X2  

≥ 0

In the uploaded image of the graph, the shaded region in the graph is the feasible region.  

c) Consider the following corner point's (0,0), (0, 53.33), (20, 40) and (40, 0) of the feasible region from the graph  

At point (0, 0), the objective function,  

500X1 + 400X2 = 500 × 0 + 400 × 0  

= 0

At point (0, 53.33), the value of objective function,

500X1 + 400X2 = 500 × 0 + 400 × 53.33 = 21332  

At point (40, 0), the value of objective function,  

500X1 + 400X2 = 500 × 40 + 400 × 0 = 20000  

At point (20, 40), the value of objective function

500X1 + 400X2 = 500 × 20 + 400 × 40 = 26000  

The maximum value of the objective function is  

26000 at corner point ( 20, 40 )

Hence, the optimal solution of this problem is  

X1 = 20, X2 = 40 and the objective is 26000

Need Help Fast!!!!!!!!!!!!!!!!!!

Answers

7.2, 3, 8.09, 2.22, 5.06, 2.5

1 poitWhat is the equation of the line in slope-intercept form of the line passing
through (1, 4) and (6, -1)? *
O y = x + 5
O y = -x + 5
y = x-5
O y = -x-5
Send me a copy of my responses.

Answers

Answer:

Its y= -x+5

Step-by-step explanation:

see image

For a field trip 9 students rode in cars andthe rest filled eight buses. How many
students were in each bus if 265 students
were on the trip?

Answers

Answer:

32 students were on each bus.

Step-by-step explanation:

First deduct 9 students from the total number: 265 - 9 = 256

Then divide the students among the total number of buses, which is 8: 256/8 = 32

So there are 32 students on each bus.

Hope this helps

Please give brainliest

Please help! 35.1 x 8.4 = ??

Answers

Answer:

294.84

should be correct, I used a calculator.

The answer is 294.84

Happy I helped.

A cocktail nut mix should have the following minimum requirements in a can of 1 pound being sold at $3.99 a can at retail stores. The following information is given: Nut Type Minimum Amt Maximum Amt Cost per pound Almond 10% 100% $2.75 Peanut 0% 50% $0.55 Walnut 20% 100% $1.70 Cashew 0% 40% $1.20 The can costs $0.10. Find the proportion of these nuts by weight to maximize the profit of the nut manufacturer. Use those proportions to select the answer closest to the profit that can be made per can if it is sold to the retail store at $3.00 a can. Group of answer choices $1.65 $1.35 $2.76 $1.77

Answers

By maximizing the amount of cheaper nuts and minimizing the amount of expensive nuts, the manufacturer should use a mix of 50% peanuts, 40% cashews, and 10% almonds. The profit per can will be $1.87, closest to $1.77 out of the given choices.

This problem involves linear programming, and it is a problem of maximizing profit under given constraints. The proportions of nuts can be found using optimization techniques that are usually covered in calculus or advanced algebra classes, but a quick answer can be given here by considering the cost of each type of nut.

The manufacturer should maximize the amount of the cheapest nuts in the mix to increase the profit. So, they should fill the can with 50% peanuts ($0.55 per lb), 40% cashews ($1.20 per lb), and 10% almonds ($2.75 per lb) to meet the minimum requirement for almonds and maximum constraints for peanuts and cashews. The average cost of this mix per pound will be (0.5x0.55)+(0.4x1.2)+(0.1x2.75) which equals $1.025 per lb. If the can costs $0.10, the total manufacturing cost per can becomes $1.125.

If the nut mix is sold at $3.00, the profit per can will be $3.00 - $1.125 which equals $1.87. This is closer to $1.77. Therefore, the answer should be $1.77 if we consider rounding to the closest number. Note that we can't fill a full can because the total proportion is already 100%, but we have ignored the volume taken up by the can itself in this calculation.

Learn more about Linear Programming here:

brainly.com/question/24361247

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Answer:

Nut

Step-by-step explanation: