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Gelneren [198K]
2 years ago
12

What's the area of the above figure which has sides of length 6 and 8

Mathematics
2 answers:
Sunny_sXe [5.5K]2 years ago
6 0
Area of a rectangle is length times width. In the figure above, the length is 8 and the width is 6. So you just multiply 8 by 6 to get 48.

A = l x w
A = 8 x 6
A = 48
san4es73 [151]2 years ago
3 0
Area means times so wouldn’t it just be 6 times 8 which is 48?!
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Step-by-step explanation:

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A population has a standard deviation of 80. A random sample of 400 items from this population is selected. The sample mean is d
Helga [31]

Answer:

ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232

Step-by-step explanation:

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =80 represent the population standard deviation

n=400 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)

Since the Confidence is 0.94 or 94%, the value of \alpha=0.06 and \alpha/2 =0.03, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that z_{\alpha/2}=1.8808

The margin of error is given by:

ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232

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Discuss the validity of the following statement. If the statement is always​ true, explain why. If​ not, give a counterexample.
Zarrin [17]

Correction:

Because F is not present in the statement, instead of working on​P(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

So

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

8 0
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