Find the slope and the y-intercept of the following linear equation. x = 4y – 16

Mathematics

1 answer:

Savatey [412]4 years ago

7 0

I hope that this picture of the math problem helps

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Min mows lawns to earn money. The points in this

Sedbober [7]

A 6hours is the answers

4 0

3 years ago

10p - (3p - 4) = 4(p + 1) + 9

andrezito [222]

Answer:

p = 3

Step-by-step explanation:

distribute parenthesis on both sides of the equation

10p - 3p + 4 = 4p + 4 + 9 ( simplify both sides )

7p + 4 = 4p + 13 ( subtract 4p from both sides )

3p + 4 = 13 ( subtract 4 from both sides )

3p = 9 ( divide both sides by 3 )

p = 3

6 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

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