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nekit [7.7K]
3 years ago
7

Rewrite

7D" id="TexFormula1" title="\frac{2}{x^{2}-x-12 } and \frac{1}{x^{2}-16 }" alt="\frac{2}{x^{2}-x-12 } and \frac{1}{x^{2}-16 }" align="absmiddle" class="latex-formula"> as equivalent rational expressions with the lowest common denominator.
Provide your answer below:
___ , ___
Mathematics
2 answers:
VashaNatasha [74]3 years ago
8 0
Dhduudifufjfhxh h h hutch u u u ucuchch hcuuchchccuucucucucucucucucucucucuc
gogolik [260]3 years ago
4 0

Answer:

2(x+4)/(x+3)(x-4)(x+4) and 1(x+3)/(x+3)(x-4)(x+4)

Step-by-step explanation:

Factor x^2-x-12: (x+3)(x-4)

Factor x^2-16: (x+4)(x-4)

So, LCD would be (x+3)(x-4)(x+4)

Rewritten: 2(x+4)/(x+3)(x-4)(x+4) and 1(x+3)/(x+3)(x-4)(x+4)

I hope this helped and have a good rest of your day!

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A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq
Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

s=16.6 represent the sample deviation

\bar X=12.8 represent the sample mean

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Confidence =99% or 0.99

\alpha=1-0.99=0.01 represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448

12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

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