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Firdavs [7]
3 years ago
10

Is 12/50 Grader than 4/5

Mathematics
1 answer:
Oliga [24]3 years ago
7 0

Answer:

No, \frac{4}{5} is greater than \frac{12}{50}.

Step-by-step explanation:

The two fractions to be compared are given as:

\frac{12}{50} and  \frac{4}{5}

Now, in order to compare them, we need to make the denominator same for both the fractions and then compare their numerators.

LCD of 50 and 5 is 100. So, we try to make each denominator 100 and change the numerator accordingly.

We multiply the first fraction by \frac{2}{2} and we multiply the second fraction by \frac{20}{20} as given below:

\frac{12\times 2}{50\times 2}=\frac{24}{100}\\\frac{4\times 20}{5\times 20}=\frac{80}{100}

Now, the two new fractions are \frac{24}{100} and \frac{80}{100}.

Now, we check the numerators. The numerator that is larger among the two has the greater fraction value.

Here, 80 is greater than 24. So, fraction \frac{80}{100} is greater than \frac{24}{100} or \frac{4}{5} is greater than \frac{12}{50}.

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Geneva is -3 °C. Nikki's hometown will be 12° warmer, or +9 °C. The city in the table with that temperature is San Franscisco.

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Please help! A clear explanation on how to simplify the exponents and a detailed step by step solution is greatly appreciated.
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What is absolute deviation?​
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The average absolute deviation (or mean absolute deviation ( MAD )) about any certain point (or 'avg. absolute deviation' only) of a data set is the average of the absolute deviations or the positive difference of the given data and that certain value (generally central values). answer:

Step-by-step explanation:

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Answer:

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12.8

The sample deviation calculated s=4.324 \approx 4.3

12.8-4.604\frac{4.324}{\sqrt{5}}=3.896    

12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704    

So on this case the 99% confidence interval would be given by (3.896;21.704)    

Step-by-step explanation:

Data: 10,9,20,13,12

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)

s represent the sample standard deviation

n 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 mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12.8

The sample deviation calculated s=4.324

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

df=n-1=5-1=4

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,4)".And we see that t_{\alpha/2}=4.604

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

12.8-4.604\frac{4.324}{\sqrt{5}}=3.896    

12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704    

So on this case the 99% confidence interval would be given by (3.896;21.704)    

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