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vlada-n [284]
3 years ago
5

Which of the following errors can be reduced only by increasing the sample size? a. systematic error b. random sampling error c.

response bias d. frame error e. selection error
Mathematics
1 answer:
garik1379 [7]3 years ago
4 0

Answer:

The correct answer is

b. random sampling error

Step-by-step explanation:

Sampling is a statistical method of selecting a predefined number of observations  from a larger population to determine the characteristics of and to represent the larger population.

A sampling error is a type of error in statistics that is as  a result of a difference between the calculated statistical values of a selected sample and that of the entire population. A sample of a population with an error does not properly represent the statistics of the population.

By the definition of sampling in statistics, sampling error can be eliminated by Increasing the size of the sample taken from the larger population, analogous central limit theorem.

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4. Given the data below, where X is the independent variable and Y is the dependent variable. If you
oksian1 [2.3K]
The answer is A, look at the difference between 0 and 15 (15) and the difference between 5 and 10 (5) then divide them 5/15.
3 0
3 years ago
Complete the following exercises by applying polynomial identities to complex numbers. Factor x2 + 64. Check your work. Factor 1
Mumz [18]
x^2+64=x^2-(8i)^2=(x-8i)(x+8i)

16x^2+49=(4x)^2-(7i)^2=(4x-7i)(4x+7i)

(x+9i)^2=x^2+2x\cdot9i+(9i)^2=x^2+18xi+81i^2=x^2+18xi-81=x^2-81+18xi

(x-2i)^2=x^2-2x\cdot2i+(2i)^2=x^2-4xi+4i^2=x^2-4xi-4=x^2-4-4xi

(x+(3+5i))^2=x^2+2x(3+5i)+(3+5i)^2\\\\=x^2+6x+10xi+3^2+2\cdot3\cdot5i+(5i)^2\\\\=x^2+6x+10xi+9+30i+25i^2=x^2+6x+10xi+9+30i-25\\\\=x^2+6x+10xi+30i-16=x^2+6x-16+(10x+30)i


Used:\\i^2=-1\\\\a^2-b^2=(a-b)(a+b)\\\\(a+b)^2=a^2+2ab+b^2
4 0
3 years ago
Read 2 more answers
Determine the number of solutions for the equation shown below.
Stolb23 [73]

Answer:

A. 0

Step-by-step explanation:

Nothing can be equal to itself if you subtract 9 from it.

6 0
3 years ago
24% of 289 estimated using<br> a rate per 100
zaharov [31]

Answer:

The answer would be 40!

Step-by-step explanation:

Estimating ~ rounding

I would round 24% (20%), and 289 (300).

using the formula:

is/of \\ = %/100,

I'm looking for "IS"

I multiply 20 (just 20, no percent) by 200 = 4000.

Then I divide by 100 to get the "IS".

That's how I got my answer!

Hope this helps!

6 0
3 years ago
Please help me with the worksheet
aivan3 [116]

Answer:

9)    a = ¾, <u>vertex</u>: (-4, 2),  <u>Equation</u>: y = ¾|x + 4| + 2

10)  a = ¼, <u>vertex</u>: (0, -3),  <u>Equation</u>: y = ¼|x - 0| - 3

11)   a = -4,  <u>vertex</u>: (3,  1),   <u>Equation</u>: y = -4|x - 3| + 1

12)  a = 1,    <u>vertex</u>: (-2, -2),  <u>Equation</u>: y = |x + 2| - 2

Step-by-step explanation:

<h3><u>Note:</u></h3>

I could <u><em>only</em></u> work on questions 9, 10, 11, 12 in accordance with Brainly's rules. Nevertheless, the techniques demonstrated in this post applies to all of the given problems in your worksheet.

<h2><u>Definitions:</u></h2>

The given set of graphs are examples of absolute value functions. The <u>general form</u> of absolute value functions is: y = a|x – h| + k, where:

|a|  = determines the vertical stretch or compression factor (wideness or narrowness of the graph).

(h, k) = vertex of the function

x = h represents the axis of symmetry.

<h2><u>Solutions:</u></h2><h3>Question 9)  ⇒ Vertex: (-4, 2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (0, 5) and substitute into the general form (equation):

y = a|x – h| + k

5 = a| 0 - (-4)| + 2

5 = a| 0 + 4 | + 2

5 = a|4| + 2

5 = 4a + 2

Subtract 2 from both sides:

5 - 2 = 4a + 2 - 2

3 = 4a

Divide both sides by 4 to solve for <em>a</em>:

\LARGE\mathsf{\frac{3}{4}\:=\:\frac{4a}{4}}

a = ¾

Therefore, given the value of a = ¾, and the vertex, (-4, 2), then the equation of the absolute value function is:

<u>Equation</u>:  y = ¾|x + 4| + 2

<h3>Question 10)  ⇒ Vertex: (0, -3)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -2) and substitute into the general form (equation):

y = a|x – h| + k

-2 = a|4 - 0| -3

-2 = a|4| - 3

-2 = 4a - 3

Add 3 to both sides:

-2 + 3 = 4a - 3 + 3

1 = 4a  

Divide both sides by 4 to solve for <em>a</em>:

\LARGE\mathsf{\frac{1}{4}\:=\:\frac{4a}{4}}

a = ¼

Therefore, given the value of a = ¼, and the vertex, (0, -3), then the equation of the absolute value function is:

<u>Equation</u>:  y = ¼|x - 0| - 3

<h3>Question 11)  ⇒ Vertex: (3, 1)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -3) and substitute into the general form (equation):

y = a|x – h| + k

-3 = a|4 - 3| + 1

-3 = a|1| + 1

-3 = a + 1

Subtract 1 from both sides to isolate <em>a</em>:

-3 - 1 = a + 1 - 1

a = -4

Therefore, given the value of a = -4, and the vertex, (3, 1), then the equation of the absolute value function is:

<u>Equation</u>:  y = -4|x - 3| + 1

<h3>Question 12)  ⇒ Vertex: (-2, -2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (-4, 0) and substitute into the general form (equation):

y = a|x – h| + k

0 = a|-4 - (-2)| - 2

0 = a|-4 + 2| - 2

0 = a|-2| - 2

0 = 2a - 2

Add 2 to both sides:

0 + 2  = 2a - 2 + 2

2 = 2a

Divide both sides by 2 to solve for <em>a</em>:

\LARGE\mathsf{\frac{2}{2}\:=\:\frac{2a}{2}}

a = 1

Therefore, given the value of a = -1, and the vertex, (-2, -2), then the equation of the absolute value function is:

<u>Equation</u>:  y = |x + 2| - 2

5 0
2 years ago
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