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miskamm [114]
3 years ago
15

5/x - 3/4 = Simplify

Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer:

\frac{20-3x}{4x}

Step-by-step explanation:

-This is an LCM problem.

-To simplify, we introduce a least common multiplier which is equivalent the product of the denominators:

LCM=x\times 4\\\\=4x

#We introduce the LCM and adjust the fractions based on it :

=\frac{5}{x}+\frac{3}{4}\\\\\\=\frac{20}{4x}+\frac{3x}{4x}\\\\\\=\frac{20-3x}{4x}

Hence, the simplified form of the fraction is: \frac{20-3x}{4x}

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The 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

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s=4 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

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A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

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

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On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=36.191

\chi^2_{1- \alpha/2}=7.633

And replacing into the formula for the interval we got:

\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}

8.400 \leq \sigma^2 \leq 39.827

So the 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

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