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____ [38]
2 years ago
8

25x2 + 36 25x2 - 36 = 25x2 + 60x + 36=​

Mathematics
1 answer:
Citrus2011 [14]2 years ago
5 0
Can you be more specific
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Which number sets (i.e. Natural, Whole, Integer, Rational, Real, etc.) are best suited to represent discrete functions? Why? Whi
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4 0
3 years ago
The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<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.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0
2 years ago
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