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vaieri [72.5K]
3 years ago
15

This is due in 15 minutes pls help me lol thx

Mathematics
2 answers:
GaryK [48]3 years ago
7 0

Answer:

ITS A

Step-by-step explanation:

PLZ GIVE BRAINLEIST

goldfiish [28.3K]3 years ago
6 0
2/3 i think, hope i answered in time, gl
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The second and fourth answer are correct

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Weite the number names <br> 31,19,624<br> 4,06,85,012<br> 6,500,000<br> 25,430,756
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Thirty-one million, six hundred and twenty-four

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A male lion eats about 108 pounds of meat in a week about how much does it eat in a day
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A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th
Valentin [98]

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)    

6 0
3 years ago
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Korvikt [17]
3/4 - 1/4 = 2/4 = 1/2

5/8 - 1/8 = 4/8 = 1/2
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