Suppose n=6 , m=2 , z1 is the average of the elements of x , and z2 is the average of the first three elements of x minus the average of fourth through sixth elements of x . Determine A . Note: Enter A in a list format: [[A11,...,A16],[A21,...,A26]]

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Answer:

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Step-by-step explanation:

Given that;

n=6 , m=2 , z1 is the average of the elements of x , and z2 is the average of the first three elements of x minus the average of fourth through sixth elements of x

So, Let d=n∧28, d′=n∧22.

We have nd=10, nd′=20n

so

n=10d=20d′nwhenced=2d′.

On the other hand, d is a divisor of 28, and above shows d′ is, too. As it is also a divisor of 22, the only possibilities are d′=1, d′=2, corresponding to d=2, d=4.

However, if d=2, n=20, and 20∧28=4, not 2. So the only solution is d=4, and n=40.

The solution is d=4 and n=40.

See step by step explanations fir answers.

n=6, m=2, z1 is the average of the elements of x, and z2 is the average of the first three elements of x minus the average of the fourth through sixth elements of x.

What is the average of the element?

Average is the sum of array elements divided by the number of elements.

So, Let d=n∧28, d′=n∧22.

We have nd=10, nd′=20n

so n=10 d=20 d′nw henced=2d′.

On the other hand, d is a divisor of 28, and the above shows, that d is, too. As it is also a divisor of 22, the only possibilities are d′=1, d′=2, corresponding to d=2, d=4.

However, if d=2, n=20, and 20∧28=4, not 2.

So the only solution is d=4 and n=40.

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