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pshichka [43]
3 years ago
10

After analyzing a data set using the one-way ANOVA model, the same data are analyzed using the randomized block design ANOVA mod

el. SS (Treatment) in the one-way ANOVA model is ________ the SS (Treatment) in the randomized block design ANOVA model.
a. Always equal to.
b. Always greater than.
c. Always less than.
d. Sometimes greater than.
Mathematics
1 answer:
Contact [7]3 years ago
4 0

Answer: Always equal to

Step-by-step explanation:

A one way analysis of variance refers to the technique that is used in knowing if there's significant difference between two samples means.

Based on the options given, it should be noted that SS (Treatment) in the one-way ANOVA model is always equal to the SS (Treatment) in the randomized block design ANOVA model.

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How do you divide 1.25 by 24
GREYUIT [131]

Answer:


Step-by-step explanation:

The awnser to this problem is +19.20 because since 24 is bigger than 1.25 it goes on the inside of the house and 1.25 goes on the outside because its smaller.

7 0
3 years ago
The endpoints of the diameter of a circle are (-12, 7) and (3, 15). What is the length of the diameter?
seraphim [82]

Answer: 17

Step-by-step explanation:

17

5 0
2 years ago
If X = -6 and Y = 10 then find |x|-|3y|/|xy|
jek_recluse [69]

Answer:

(2)

Step-by-step explanation:

6 0
3 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

5 0
3 years ago
Joe plotted the points (2, 0.5), (1, 1), and (0.5, 2), and claimed that these points represent an inverse variation relationship
Kobotan [32]
Xy = 1 for all points.

 We are going to demonstrate this affirmation:
 (2) * (0.5) = 1
 (1) * (1) = 1
 (0.5) * (2) = 1
 Therefore, the relationship between the ordered pairs is inverse.
 The relationship between the ordered pairs is:
 y = 1 / x

 Answer: 
 xy = 1 for all points.
7 0
3 years ago
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