⦁ Solve for x: 3|x – 7| = 15 SHOW WORK

Mathematics

2 answers:

nasty-shy [4]2 years ago

5 0

Ez way to do it use distribution 3x-21=15 add 21 to 15 =36
3x=36 divided by 3 both side x=12 x=2

Anna007 [38]2 years ago

4 0

$3|x-7| = 15\\\\\implies |x-7| =\dfrac{15}3\\\\\implies |x -7| = 5\\\\\implies x -7 = 5 ~~ \text{or}~ x - 7 = -5\\\\\implies x = 12 ~~\text{or}~~ x = 2$

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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

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