Help please I just need this

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

Contact [7]

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.

4 0

2 years ago

Does any one know how to do this. It would be such a life saver if you could help me out. Thx

Aneli [31]

Answer:

1: (3 ,- 4)

2:up (a > o)

3: -4

4: 2 units to the right

5: 4 units down

6: y = (x-2)^2-4

7:

8:

9: (5,9)

10: x=5

11: (4,7) and (6,7)

12:

Step-by-step explanation:

I'm not sure if you wanted it explained or answered , so i just answered most of them. Sorry if you wanted it explained.

3 0

2 years ago

Jennifer needs to make an average of at least $350 per week. If she makes $385, $326, and $298 for the first 3 weeks of the mont

ella [17]

385 + 326 + 298 = 1,009 / 3 = 336.33

385 + 326 + 298 + x = 350

350 x 4 = 1,400
1,400 - 1,009 = 391

385 + 326 + 298 + 391 = 1,400 / 4 = 350

The answer is $391

8 0

3 years ago

Which ordered pair is a solution of the equation 2x-3y=19

goldenfox [79]

D because 5 times 2 is 10 and - cuts off - so -3 times -3 would be 9 so 10+9 is 19 :)

7 0

2 years ago

Read 2 more answers

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$