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Lunna [17]
3 years ago
5

g Which statement is correct? Group of answer choices A sample size of 1 is never acceptable. Judgment sampling is preferred to

systematic sampling. Cluster sampling is useful when strata characteristics are unknown. Sampling without replacement introduces bias in our estimates of parameters.
Mathematics
1 answer:
Oxana [17]3 years ago
7 0

Answer:

A sample size of 1 is never acceptable

Explanation:

A sample size of 1 is simply not statistically significant enough as to reach close to accurate conclusions based on sample data. Statistical tests using just one data point may not be enough to test data variability. Axrule of thumb or best practice is to have at least 10% sample size from the population. It is advised to have at least a sample size of 100 sayvpopulation is 1000 to be able to reach more valid conclusions from sample data.

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3 years ago
I want to do this problem in product standard form (5i-3)(2i+1)
Ilia_Sergeevich [38]
<h3>Answer:    -13 - i</h3>

=======================================

Work Shown:

Let x = 5i-3

That allows us to go from (5i-3)(2i+1) to x(2i+1)

Distribute the x through: x*(2i) + x*(1) = 2i*x + x = 2i(x) + 1(x)

Now we replace every x in 2i(x) + 1(x) with 5i-3, and then we distribute a second time

2i(x) + 1(x) = 2i(5i-3) + 1(5i-3)

2i(x) + 1(x) = 2i(5i)+2i(-3) + 1(5i)+1(-3)

2i(x) + 1(x) = 10i^2 - 6i + 5i - 3

2i(x) + 1(x) = 10(-1) - i - 3

2i(x) + 1(x) = -10 - i - 3

2i(x) + 1(x) = -13 - i

Therefore, (5i-3)(2i+1) = -13 - i

The result is in a+bi form where a = -13 and b = -1.

------------------

An alternative method is to use the box method. This is where you set up a grid that helps you multiply out (5i-3)(2i+1)

See the diagram below.

Each of the 4 red terms in the boxes represents the result of multiplying the outer blue and green terms. Example: 5i times 2i = 10i^2 in row1, column1.

7 0
3 years ago
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