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xeze [42]
3 years ago
10

At a college bookstore, Tanisha bought a Math Textbook and a Novel for $54. If the price of the Math Textbook is $8 more than 3

times the price of the Novel, then how much is the price of each book?
Mathematics
1 answer:
pishuonlain [190]3 years ago
3 0

Answer: Price of Math Textbook= $42.5,  Price of the Novel= $11.5

Step-by-step explanation:

Let x = price of the Novel.

Price of Math Textbook = 8+3x

As per given , we have

Price of Math Textbook + Price of the Novel = $54

⇒ 8+3x+x = 54

⇒ 8+4x=54

⇒ 4x = 46     [subtract 8 from both sides]

⇒ x = 11.5

Price of Math Textbook = 8+3(11.5) = 8+34.5= $42.5

Hence, Price of Math Textbook= $42.5,  Price of the Novel= $11.5

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Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of \mu = 1.35.
  • The standard deviation is of \sigma = 0.33.

Item 1:

Considering his ratio, we have that X = 0.73, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.73 - 1.35}{0.33}

Z = -1.88

His z-score was of Z = -1.88.

Item 2:

The probability is the <u>p-value of Z = -1.88</u>, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{1.96 - 1.35}{0.33}

Z = 1.85

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213

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