Here are some questions I'm confused about

Mathematics

2 answers:

Ann [662]3 years ago

6 0

16. -0.65, -3/8, 5/16, and 2/4

18. -5/8, -0.6, -7/12, 0.72

20. -0.35, -0.3, 0.5, 0.55

Hope this helps!.

Neporo4naja [7]3 years ago

5 0

16. -0.65 , -3/8 , 5/16 , 2/4

18. -7/12 , -0.6 , -5/8 , 0.72

20. -0.35 , 0.3 , 0.5 , 0.55

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What number multiplied by itself 3 times equals-343

professor190 [17]

343*3=1029 this is the answer

7 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

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