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iragen [17]
3 years ago
8

A farmer can buy two types of plant​ food, mix A and mix B. Each cubic yard of mix A contains 35 pounds of phosphoric​ acid, 15

pounds of​ nitrogen, and 16 pounds of potash. Each cubic yard of mix B contains 14 pounds of phosphoric​ acid, 18 pounds of​ nitrogen, and 64 pounds of potash. The minimum monthly requirements are 840 pounds of phosphoric​ acid, 540 pounds of​ nitrogen, and 768 pounds of potash. If x is the number of cubic yards of mix A used and y is the number of cubic yards of mix B​ used, write a system of linear inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the amounts of mix A and mix B that can be used.

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
7 0

Answer:

The answer is below

Step-by-step explanation:

Given that the minimum monthly requirements are 840 pounds of phosphoric​ acid, hence:

35 pounds of phosphoric acid mix A + 14 pounds of phosphoric acid mix B ≥ 840 pounds of phosphoric​ acid

35x + 14y ≥ 840

The minimum monthly requirements are 540 pounds of nitrogen, hence:

15 pounds of nitrogen mix A + 18 pounds of nitrogen mix B ≥ 540 pounds of nitrogen

15x + 18y ≥ 540

The minimum monthly requirements are 768 pounds of potash, hence:

16 pounds of potash mix A + 64 pounds of potash mix B ≥ 768 pounds of potash

16x + 64y ≥ 768

Also, x ≥ 0 and y ≥0.

Therefore the equations are:

35x + 14y ≥ 840

15x + 18y ≥ 540

16x + 64y ≥ 768

x ≥ 0 and y ≥ 0

Plotting with geogebra gives the graph below.

The point are (0,60), (48,0), (18,15), (32,4)

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Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So we use the binomial probability distribution to solve this question.

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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(a) What is the probability that more than 7 preferred McDonald's?

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

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0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

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