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qwelly [4]
2 years ago
12

Marine fuel is a combination of gasoline and motor oil. The standard gasoline and oil mixture is about 98% gasoline and about 2%

motor oil. The "break in" mixture for a new engine is about 96% gasoline and about 4% motor oil. a. Write an equation in standard form that models the possible combinations of each kind of mixture you can prepare using 6 gallons of gasoline. b. If you prepare 4 gallons of the "break in" mixture, how much gasoline will you have for the standard mixture?
Mathematics
1 answer:
Georgia [21]2 years ago
6 0

Answer:

0.98g+ 0.02m= 7.1224 gallons standard mixture

0.96g+ 0.04m= 6.25 gallons break oil

3.92 gallons of gasoline  is required for  4 gallons the standard mixture

Step-by-step explanation:

If we have to use 6 gallons of gasoline this means that 98%= 6 gallons and 2% would be 6/98*2

Gasoline: Motor oil

6              :    x

98        : 2

Using the product rule

x= 6*2/98= 0.1224 gallons of motor oil is required.

Let the gasoline be represented by g and motor oil by m then for the standard oil the equation would be

0.98g+ 0.02m= 7.1224 gallons standard mixture

For break in oil

Gasoline: Motor oil

6              :    x

96        : 4

Using the product rule

x= 6*4/96= 0.25 gallons of motor oil is required.

Let the gasoline be represented by g and motor oil by m then for the break oil the equation would be

0.96g+ 0.04m= 6.25 gallons break oil

Part b. For  4 gallons of the "break in" mixture

96% of the 4 gallons of the "break in" mixture is gasoline =  3.84 gallons of gasoline

4% of 4 = 0.16 gallons of motor oil

Now the standard mixture 4 gallons would contain 98 % of gasoline and 2 % of motor oil

98% of 4 gallons= 3.92 gallons of gasoline

2% of  4 gallons= 0.08 gallons of motor oil.

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Slope of a horizontal line. When two points have the same y-value, it means they lie on a horizontal line. The slope of such a line is 0, and you will also find this by using the slope formula.

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Step-by-step explanation:

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In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
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Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

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