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Sonja [21]
3 years ago
11

Delaney would like to make a 5 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of peanuts and several

pounds of a mixture that is 20% peanuts and 80% almonds. Let p represent the number of pounds of peanuts needed to make the new mixture, and let m represent the number of pounds of the 20% peanuts-80% almonds mixture.
(a) What is the system that models this situation? (8 points)
(b) Which of the following is a solution to the system
2 lb peanuts and 3 lb mixture;
2.5 lb peanuts and 2.5 lb mixture;
4 lb peanuts and 1 lb mixture? Show your work. (7 points)
ONLY ANSWER WITH WORK AND ANSWER IF YOU KNOW, MAKE IT EASY TO UNDERSTAND BTW 20 POINTS
Mathematics
1 answer:
Alchen [17]3 years ago
7 0

Answer:

a) The system that models the situation is:

p + 0.2m = 3

p + m = 5

b) The solution of the system is:

m = 2.5

p = 2.5

Step-by-step explanat:

It is desired to obtain 5 pounds of a mixture containing 40% peanuts and 60% almonds.

It has a mixture of 20% peanuts and 80% almonds.

p: number of pounds of peanuts

m = amount of mixture

The equation that represents the amount of peanut and mixture that is needed is:

p + m = 5 pounds

m = 5-p (i)

The amount of peanut in total that the mixture should have is:

p + 0.2m = 0.6 (5)

p + 0.2m = 3 (ii)

a) The system that models the situation is:

p + 0.2m = 3

p + m = 5

Now we substitute (i) in (ii) and clear p.

p + 0.2 (5-p) = 3

p-0.2p = 2

p = 2.5 pounds

Now we find m

m = 5 - 2.5

m = 2.5 pounds

b) The solution of the system is:

m = 2.5

p = 2.5

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Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

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Binomial probability distribution

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

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that p = 0.15

Sample of 6:

This means that n = 6

What is the probability that there will be 0 or 1 defects in a sample of 6?

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In which

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P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764

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