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Murljashka [212]
3 years ago
5

Please answer quick

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
4 0

Answer:

9,850

Step-by-step explanation:

Sunny_sXe [5.5K]3 years ago
4 0

Answer:

9850

Step-by-step explanation:

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What is the volume of a cylinder whose radius is 5 and height is 9
liraira [26]

Answer:

225π

Step-by-step explanation:

substitute V=πr^2h

V=π(5)^2(9)

Simplify.

225π

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3 years ago
Which of the following is the correct conversion for 1,200cm?
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The answer is 2, 12 meters
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each car on a commuter train can seat 114 passengers. if the train has 7 cars how many passengers can the train seat?
Sav [38]
144 x 7 = 1008

The train can hold 1008 people
6 0
3 years ago
Read 2 more answers
43) Kristin is carrying nine pages of math homework and four pages of English homework. A gust of wind blows the pages out of he
Aleksandr-060686 [28]

Answer: 10 pages

Step-by-step explanation:

The 1st is 6/10

Then 5/9, 4/8, 3/7, 2/6, 1/5

-----

It's the product:

= 6/10, 5/9, 4/8, 3/7, 2/6, 1/5

= 3/5, 5/9, 1/2, 3/7, 1/3, 1/5

= 1/210

8 0
3 years ago
When an ANOVA comparing the means of 3 groups indicates that at least one group is different from the others, a common follow-up
lorasvet [3.4K]

Answer:

ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C

Since we reject the null hypothesis we want to see which method it's the best to determine which group(s) is (are) different is pairwise two-sample t-tests each assessed using.

And on this case the best option is:

ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

The reason is because the Bonferroni correction "compensates for that increase by testing each individual hypothesis at a significance level of \alpha who represent the desired overall alpha level and m is the number of hypotheses". For our case m=3 hypotheses with a desired \alpha = 0.05, then the Bonferroni correction would test each individual hypothesis at 0.05/3=0.0167

One advatange of this method is that "This method not require any assumptions about dependence among the p-values or about how many of the null hypotheses are true" . And is more powerful than the individual paired t tests.

3 0
4 years ago
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