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navik [9.2K]
3 years ago
6

How many solutions does this equation have

Mathematics
1 answer:
Stells [14]3 years ago
8 0

Answer:

I m not sure, I can't see

Step-by-step explanation:

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The quality-control manager of a large factory is concerned about the number of defective items produced by workers. Thirty work
kotegsom [21]

Answer:

a. There is no blocking variable, and incentive plans will be randomly assigned to the workers.

Step-by-step explanation:

The Randomized Complete Block Design (RCBD) is a standard experimental design where experimental units or subjects are grouped as blocks (also known as replicates). In RCBD, subjects within each block are randomly assigned to the experimental units within a block. RCBD is a type of design that reduces variability by controlling variation within each treatment, thereby enhancing the estimation of the treatment effects (combinations of the factor levels of the different factors).

6 0
3 years ago
What is the product of 4xy and y2 + 2x?
juin [17]
The product of 4xy and y² + 2x can also be written as
4xy(y² + 2x)

When we have brackets like this, we multiply whatever is left of the brackets by everything inside.

1) 4xy multiplied by y²
    4xy x y² = 4xy³

2) 4xy multiplied by 2x
    4xy x 2x = 8x²y

3) Add together 4xy³ and 8x²y 
    4xy³ + 8x²y
3 0
3 years ago
The product of 4 and a number x is 14
Sever21 [200]

Answer:

x=3.5

Why?

14/4=3.5

3 0
3 years ago
(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0
3 years ago
Pleaseeeeeeeeeee helpp​
lord [1]

Answer:

0

9

18

27

Step-by-step explanation:

# of cards * 1.8

4 0
2 years ago
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