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Alexandra [31]
2 years ago
10

A 25.5-g piece of cheddar cheese contains 37% fat, 28% protein, and 4% carbohydrate. the respective fuel values for protein, fat

, and carbohydrate are 17, 38, and 17 kj/g, respectively. the fuel value for this piece of cheese is ________ kj.
Mathematics
1 answer:
marta [7]2 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

m(fat) = %fat x 25,5 = 0,37 x 25,5 = 9,435 g of fat 

<span>m(protein) = %protein x 25,5 = 0,28 x 25,5 = 7,14 g of protein </span>

<span>m(carbohydrate) = %carbohydrate x 25,5 = 0,04 x 25,5 = 1,02 g of carbohydrate </span>



<span>E = m(protein) x 17 + m(fat) x 38 + m(carbohydrate) x 17 </span>
<span>E = (7,14 x 17) + (9,435 x 38) + (1,02 x 17) </span>
<span>E = 497 kJ</span>
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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is   p =\frac{1}{5}

The correct answer choice is

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b

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c

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d

The correct answer is

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Using the R the probability Pe = 0.0003

The probabilities P \approx Pe

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

                  p =\frac{1}{5}

                    =0.2

The mean of this distribution is mathematical represented as

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substituting the value

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             \mu = 20

The standard deviation is mathematically represented as

         \sigma = \sqrt{np (1-p)}

substituting the value

           = \sqrt{100 * 0,2 (1-0.2)}

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Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

               P=P(X \ge 34 )

By standardizing the normal approximation we have that

              P(X \ge 34) \approx P(Z \ge z)

Now z is mathematically evaluated as

               z = \frac{x-\mu}{\sigma }

Substituting values

             z = \frac{34-20}{4}

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So  using the z table the P(Z \ge 3.5) is  0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small  So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

 

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First of all, lets consider that you made a litte mistake and you meant this problem.........

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This is a system of two equations.

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