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3 years ago

Find the L.C.M of x^2+x , x^2-1 , x^2-x

Mathematics

2 answers:

Shalnov [3]3 years ago

8 0

Answer:

$x\cdot(x+1)\cdot(x-1)=x^3-x$

Step-by-step explanation:

To find the LCM of any integer, we take the product of all the <em>highest powers</em> of the factors that appear in the numbers. Factoring our three given expressions gives us the products

$x^2+x=x\cdot(x+1)\\x^2-1=(x+1)\cdot(x-1)\\x^2-x=x\cdot(x-1)\\$

Our LCM will be a product of some powers of $x$ , $x+1$ , and $x-1$ . The most each factor occurs in the three expressions is once, so our least common multiple is

$x^1\cdot(x+1)^1\cdot(x-1)^1$

which can be simplified to $x^3-x$

grin007 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

x² + x = x * (x + 1)

x² - 1 = (x - 1)(x + 1) {a² - b² = (a+b)(a-b)}

x² - x = x * (x - 1)

LCM = x * (x + 1) * (x - 1)

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butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .