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3 years ago

Solve this problem???

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

The denominator of the first term is a difference of squares, such that

4a ² - b ² = (2a)² - b ² = (2a - b) (2a + b)

So you can write the fractions as

(4a ² + b ²)/((2a - b) (2a + b)) - (2a - b)/(2a + b)

Multiply through the second fraction by 2a - b to get a common denominator:

(4a ² + b ²)/((2a - b) (2a + b)) - (2a - b)²/((2a + b) (2a - b))

((4a ² + b ²) - (2a - b)²) / ((2a - b) (2a + b))

Expand the numerator:

(4a ² + b ²) - (2a - b)²

(4a ² + b ²) - (4a ² - 4ab + b ²)

4ab

So the original expression reduces to

4ab / ((2a - b) (2a + b))

4ab / (4a ² - b ²)

upon condensing the denominator again.

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Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

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 means 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.

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".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$