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Nina [5.8K]
3 years ago
7

A random sample was taken to determine whether students from a certain classroom prefer to shop at Store A, Store B, or Store C.

Which of the following would represent a random sample?
selecting all the students that have a last name that begins with R
selecting all the girls
putting all the names in a hat and selecting six students
putting all the names in alphabetical order and selecting every fourth student
Mathematics
2 answers:
steposvetlana [31]3 years ago
8 0
Putting all the names in a hat and selecting six students because the rest would not have been random because you know who you picked just by looking but if u put names in a hat you don't know who you picked.
g100num [7]3 years ago
5 0
Putting the names in a hat and selecting six students
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Which rational number equals 0 point 4 with bar over 4?
Ulleksa [173]
<h3>Answer:   4/9</h3>

==============================================

Work Shown:

x = 0.\overline{4}\\\\x = 0.4444\ldots\\\\10x = 4.4444\ldots\\\\10x-x = 4.4444\ldots-0.4444\ldots\\\\9x = 4\\\\x = \frac{4}{9}\\\\\frac{4}{9} = 0.\overline{4}= 0.4444\ldots\\\\

The idea is that when we subtract 10x and x, the infinite decimal pattern (the 4s that go on forever) will cancel. That leads to 9x = 4 which solves to x = 4/9

6 0
3 years ago
In a survey, the planning value for the population proportion is p* = 0.35. How large a sample should be taken to provide a 95%
lions [1.4K]

Answer:

n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59  

And rounded up we have that n=350  

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".  

p represent the real population proportion of interest

\aht p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

z represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

And replacing into equation (b) the values from part a we got:  

n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59  

And rounded up we have that n=350  

6 0
3 years ago
Solve the right triangle. Round decimal answers to the nearest tenth.
dem82 [27]

Step-by-step explanation:

RS = 15/ tan57° = 9.7

RT = 15/sin57° = 17.9

3 0
3 years ago
5. On a recent day, the exchange rate for US dollars to British pounds
attashe74 [19]
250*0.62= 155

155 pounds is your answer.
8 0
3 years ago
6xy (1/2x ^2 - 1/2xy + 1/2y ^2 )
nevsk [136]
Is this correct on what you typed above -> \bf 6xy\left( \cfrac{1}{2x^2}-\cfrac{1}{2xy}+\cfrac{1}{2y^2} \right) ?
6 0
3 years ago
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