Answer:
The answer is below
Step-by-step explanation:
For a normal distributed population with a mean (μ), and a standard deviation (σ), if a sample size of n is selected from the population, the mean of the sample ([tex]\mu_x[/tex]) = μ and the standard deviation of the sample ([tex]\sigma_x[/tex]) = [tex]\frac{\sigma }{\sqrt{n} }[/tex]
Let the normal distribution population have a standard deviation of σ. If the standard deviation is to be decreased by half, the sample size (n) needed is:
[tex]Using:\\\\\sigma_x=\frac{\sigma}{\sqrt{n} } \\\\but\ \sigma_x=\frac{\sigma}{2}\\\\Hence:\\\\ \frac{\sigma}{2}=\frac{\sigma}{\sqrt{n} }\\\\Divide\ through\ by \ \sigma\ to\ get:\\\\ \frac{1}{2}=\frac{1}{\sqrt{n} }\\\\\sqrt{n}=2\\\\square\ both\ sides:\\\\(\sqrt{n} )^2=2^2\\\\n=4\\\\[/tex]
To decrease the standard deviation of the sampling distribution by half we need a sample size of 4
What is the number based on the list of factors? 1,3,7, __
Please help
Answer:
21 I think, might be wrong.
2+4x=2(2x+1)
Which statements are true?
Select each correct answer.
There is no value for x that will make the equation true.
The equation is an identity,
The solution is 2.
There are infinitely many solutions.
The equation is a contradiction.
There are no solutions.
The equation is true for all values of x.
Which graph represents this system? 2x-5v=-5 y=-3x+1
Answer:
Please check the attached graph.
Step-by-step explanation:
The slope-intercept form of the line equation
[tex]y = mx+b[/tex]
where
m is the slopeb is the y-interceptGiven the system of equations
2x - 5y = -5 --- [Equation 1]
y = -3x + 1 --- [Equation 2]
converting the equation 1 in the slope-intercept form
2x - 5y = -5
5y = 2x+5
divide both sides by 5
y = 2/5x + 1
Thus, the y-intercept = 1
Now, please check the attached graph below.
The red line represents the equation 2x - 5y = -5
From the graph, it is clear that the red line has the y-intercept y = 1.
Now, considerng the equation 2
y = -3x + 1
comparing with the slope-intercept form of line equation
The y-intercept b = 1
Hence, the y-intercept of the line y = -3x + 1 is y = 1
The green line represents the equation y = -3x + 1
From the graph, it is clear that the green line has the y-intercept y = 1.
Point of Intersection:
It is clear from the graph:
The red line represents the equation 2x - 5y = -5.The green line represents the equation y = -3x + 1.As both lines meet or intersect at the point (0, 1).
Therefore, the point of intersection of both the lines is:
(x, y) = (0, 1)
As we know that the point of intersection of both the lines represents the solution to the system of equations.
Therefore, (0, 1) on the graph represents the solution to the system of equations.
The graph is attached below.
I need help ASAP no rocky
Answer:
no rocky
Step-by-step explanation:
Answer:
The answer is C
Step-by-step explanation:
I think it's C and I am always right so it is C! You're welcome!
Certain pieces made by an automatic lathe are subject to three kinds of defects X, Y, Z. A sample of 1000 pieces was inspected with the following results: 2.1% had type X defect. 24% had type Y defect. 2.8% had type Z defect. 0.3% had both type X and type Y defects. 04% had both type X and type Z defects. 0.6% had both type Y and type Z defects. 0.1% had type X, type Y, and type Z defects. Draw a Venn Diagram, then find:
(a) What percent had none of these defects?
(b) What percent had at least one of these defects?
(c) What percent were free of type X and/or type Y defects?
(d) What percent had not more than one of these defects?
Answer:
a) % age of samples containing none of these defects = 93.9%
b)% age of samples containing at least one of these defects = 6.1%
c) % age of samples free of type X and/or type Y defects = 95.8%
d) %age of samples with not more than 1 defect = 98.9%
Step-by-step explanation:
Data Given:
Number of Samples = 1000
Type X defect = 2.1% = 21 samples
Type Y defect = 2.4% = 24 samples
Type Z defect = 2.8% = 28 samples
Both Type X and Y defect = 0.3% = 3 samples
Both Type X and Z defect = 0.4% = 4 samples
Both Type Y and Z defect = 0.6% = 6 samples
Type X and Y and Z defect = 0.1% = 1 sample
Venn Diagram is attached in the attachment below. Please refer to attachment for the Venn Diagram.
a) % age of samples containing none of these defects.
Solution:
Number of samples containing none of these defects = Total - Samples with defects
Number of samples containing none of these defects = 1000 - { (Type X) + (Type Y) + (Type Z) - (Both X and Y) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }
Number of samples containing none of these defects = 1000 - { (21) + (24) +(28) - (3) -(4) - (6) + (1) }
Number of samples containing none of these defects = 1000 - 61
Number of samples containing none of these defects = 939
% age of samples containing none of these defects = 939/1000 x 100
% age of samples containing none of these defects = 93.9%
b) % age of samples containing at least one of these defects:
We have already calculated this above, number of samples containing at least on of these defects:
number of samples containing at least on of these defects = { (Type X) + (Type Y) + (Type Z) - (Both X and Y) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }
number of samples containing at least on of these defects = { (21) + (24) +(28) - (3) -(4) - (6) + (1) }
number of samples containing at least on of these defects = 61
% age of samples containing at least one of these defects = 61/1000 x 100
% age of samples containing at least one of these defects = 6.1%
c) % age of samples free of type X and/or type Y defects.
For this find, we need to find the samples with only Z Type defect.
Number of Samples with Only Z type defects = { (Type Z) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }
Number of Samples with Only Z type defects = { (28) -(4) - (6) + (1) }
Number of Samples with Only Z type defects = 19
Now, we also know the number of samples without any defects = 939
Now,
The number of samples free of type X and/or type Y defect = Sum of Number of Samples with Only Z type defects and number of samples without any defects
The number of samples free of type X and/or type Y defect = 19+939
The number of samples free of type X and/or type Y defect = 958
% age of samples free of type X and/or type Y defects = 958/1000 x 100
% age of samples free of type X and/or type Y defects = 95.8%
d) %age of samples with not more than 1 defect:
For this find, we need to find number of samples with only X type and with only type Y and with only type Z.
We have already found the number of samples with only Z type defect = 19
Now,
number of samples with only X type defect = { (Type X) - (Both X and Z) - (Both X and Y) + (All defects X and Y and Z) }
number of samples with only X type defect = { (21) -(4) - (3) + (1) }
number of samples with only X type defect = 15
Similarly,
number of samples with only Y type defect = { (Type Y) - (Both Y and Z) - (Both X and Y) + (All defects X and Y and Z) }
number of samples with only Y type defect = { (24) -(6) - (3) + (1) }
number of samples with only Y type defect = 16
For,
samples with not more than 1 defect = number of samples with only Y type defect + number of samples with only X type defect + number of samples with only z type defect + number of samples without any defects
samples with not more than 1 defect = 939 + 16 + 15 + 19
samples with not more than 1 defect = 989
%age of samples with not more than 1 defect = 989/1000 x 100
%age of samples with not more than 1 defect = 98.9%