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True [87]
4 years ago
8

Place is highest to lowest 3/4 84% 1/3 0.82

Mathematics
1 answer:
sesenic [268]4 years ago
4 0

ANSWER

84%,0.82,¾,⅓

EXPLANATION

Convert everything to decimals.

\frac{3}{4}  = 0.75

84\% = 0.84

\frac{1}{3}  = 0.333....

The last one is already in decimals.

0.82

We can now see clearly, that

0.84  \: > \:  0.82 \:  >  \: 0.75  \: >  \: 0.3333...

This implies that,

84\% \: > \:  0.82 \:  >  \:  \frac{3}{4}   \: >  \:  \frac{1}{3}

Hence from highest to least, we have

84%,0.82,¾,⅓

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Answer:

Question 9: Variables: (smallest) s, q, r (largest)

Question 10: 5 whole numbers (7, 8, 9, 10, and 11)

Step-by-step explanation:

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<em />

For question 10, it states \frac{1}{4}. This can be split into \frac{3}{x} and \frac{3}{x} . When x is 12 in the first equation then \frac{3}{12} = \frac{1}{4} and when x is 6 in the second equation \frac{3}{6} =0.5 (0.5 is also  \frac{1}{2}). Therefore, x must be a whole number less than 12 and greater than 6, and it cannot be either 12 or 6. Whole numbers between 6 and 12 are 7, 8, 9, 10, and 11  or  5 whole numbers.

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3 years ago
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3 years ago
The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv
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Answer:

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

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=14.42 represent the sample mean

\mu population mean (variable of interest)

\sigma=6.95 represent the population standard deviation

n=36 represent the sample size  

Solution to the problem

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

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