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Travka [436]
3 years ago
12

How do I write each sum as a product of the gcf of 75 +90

Mathematics
2 answers:
vladimir1956 [14]3 years ago
8 0

Answer:

15(5+6)

Step-by-step explanation:

gcf of 75 +90

To find GCF lets break the numbers one by one

75 can be written as 5 times 5 times 3

90 can be written as 5 times 2 times 3 times 3

Now we look at the greatest common factor

Both numbers have 5 and 3 in common

So  GCF is 15

we take out 15 that is in common

15(5+6)

schepotkina [342]3 years ago
3 0
75=5*3*5
90=5*3*6

75+90=5*3(5+6)=5*3*11
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Answer:

The 99% confidence interval would be given by (196.2;283.8)

We are 99% condident that the true mean is between 196.2 and 283.8  

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.  

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".

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\bar X=240 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=15 represent the sample standard deviation

n=4 represent the sample size  

Part a

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

df=n-1=4-1=3

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(0.005,3)".And we see that t_{\alpha/2}=5.84

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

240-5.84\frac{15}{\sqrt{4}}=196.2    

240+5.84\frac{15}{\sqrt{4}}=283.8

So on this case the 99% confidence interval would be given by (196.2;283.8)    

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What assumption about the data is necessary for the inference derived from the analysis to be valid?

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.

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3 years ago
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