Find the sum to the equation of:

Mathematics

1 answer:

solong [7]3 years ago

5 0

$\frac{a}{a^{2} - 4a + 4} + \frac{7}{a^{2} + 2a - 8}$ = $\frac{a(a^{2} + 2a - 8)}{(a^{2} - 4a + 4)(a^{2} + 2a - 8)} + \frac{7(a^{2} - 4a + 4)}{(a^{2} - 4a + 4)(a^{2} + 2a - 8)}$ = $\frac{a^{3} + 2a^{2} - 8a}{a^{4} - 2a^{3} - 12a^{2} + 40a - 32} + \frac{7a^{2} - 28a + 28}{a^{4} - 2a^{3} - 12a^{2} + 40a - 32}$ = $\frac{a^{3} + 9a^{2} - 36a + 28}{a^{4} - 2a^{3} - 12a^{2} + 40a - 32}$

You might be interested in

Which expression below represents,”A number increased by five?”

Anna35 [415]

C
The word “increased” means to add, so C would be the correct answer. Hope this helps!

8 0

3 years ago

Read 2 more answers

Where my math wizzes at? I really need help and fast. A dialation has a center (0, 0, 0). Find the image of the point (-1, 2, 0)

defon

Multiply every coordinate by the scale factor 3

(-1,2,0) ----> (-3,6,0) 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

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