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vesna_86 [32]
3 years ago
9

What is 4/5 of 2? write answer in decimal form.?

Mathematics
1 answer:
mylen [45]3 years ago
5 0

The decimal form of 4/5 is .8, which can also be written as 0.8 or 0.80.Fractions can be converted into decimals using a calculator or by doing the math manually

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2 1/7+3/7(3X-5) =-4
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Question: Researchers in Pakistan wanted to better understand the effects of anthracycline (a chemotherapeutic drug) on the hear
Sedbober [7]

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}

  • It measures how many standard deviations the measure is from the mean.
  • 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

7 0
2 years ago
Divide 4 1/4 by 2. ​
Aleonysh [2.5K]
I’m pretty sure it’s 2 1/8
5 0
3 years ago
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