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liq [111]
2 years ago
7

What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu

s 9 x EndFraction minus StartFraction x + 1 Over x squared minus 9 EndFraction
Mathematics
2 answers:
inn [45]2 years ago
8 0

Answer:

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

Step-by-step explanation:

The given expression is

\frac{2x+5}{x^{2} -3x} -\frac{3x+5}{x^{3} -9x} -\frac{x+1}{x^{2}-9}

First, we need to factor each denominator

\frac{2x+5}{x(x-3)} -\frac{3x+5}{x(x+3)(x-3)} -\frac{x+1}{(x-3)(x+3)}

So, the least common factor (LCF) is x(x-3)(x+3), because they are the factors that repeats.

Now, we diviide the LCF by each denominator, to then multiply it by each numerator.

\frac{(x+3)(2x+5)-3x-5-x(x+1)}{x(x-3)(x+3)} =\frac{2x^{2}+5x+6x+15-3x-5-x^{2}-x }{x(x-3)(x+3)}\\\frac{x^{2}+7x+10}{x(x-3)(x+3)}

Then, we factor the numerator, to do so, we need to find two numbers which product is 10 and which sum is 7.

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

Therefore, the expression is equivalent to

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

expeople1 [14]2 years ago
7 0

Answer:

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

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Answer:

ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434

Step-by-step explanation:

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The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The confidence interval for a proportion is given by this formula  

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For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.  

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