An experiment has four groups and six observations in each group. What are the treatment degrees of freedom?a. 19b. 3c. 4d. 23e.
2 answers:
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Answer:
In First method : counting up, counting back on a number line,
If we want the quotient after dividing the number by 5 then we count how many 5 we get from 0 to the dividend.
For example :
Since, from 0 to 30 there are six 5's obtained. ( because 5 × 6 = 30 )
Thus,
In Second Method : dividing by 10, and then doubling the quotient.
First we divide the number by 10 then multiply the quotient by 2.
For Example:
Since,
Thus,
Now, when we compare the above methods then we conclude that for the smaller numbers first method is appropriate because for small numbers we can easily count total 5's from 0. While for large numbers Second method is appropriate because it is hard to count the total 5's for the large number.
Answer:
Step-by-step explanation:
y > 3x + 10
y < -3x - 1
<h3> Part A:</h3>You'd graph the given systems of linear inequalities the same way as you graph the linear equations.
To graph y > 3x + 10, plot the y-intercept, (0, 10), then use the slope, m = 3 (rise 3, run 1), to plot other points on the graph. Use a dashed line (because of the ">" symbol).
Follow the same steps for the other linear inequality. Plot the y-intercept, (0, -1), then use the slope, m = -3 (down 3, run 1) to plot other points. Use a dashed line (because of the "<" symbol).
Pick a test point on either side of the boundary line and plug it into the original problem. This will help determine which side of the boundary line is the solution. Plug in a test point that is not on the boundary line.
Use the point of origin, (0, 0) as the test point. Plug in these values into the given systems of linear inequalities to see whether it will provide a true statement.
y > 3x + 10
0 > 3(0) + 10
0 > 0 + 10
0 > 10 (False statement).
y < -3x - 1
0 < -3(0) - 1
0 < 0 - 1
0 < -1 (false statement).
Since the point of origin provided a false statement to the given systems of linear inequalities, you must shade the half-plane region where it doesn't contain the test point.
<h3>Part B: </h3>
You'll do the same process as what I've done for the test point. Plug in the values of (8, 10) into the given systems of linear inequalities. If it provides a false statement, then it means that it is not a solution to the system.
y > 3x + 10
10 > 3(8) + 10
10 > 24 + 10
10 > 34 (False statement).
y < -3x - 1
10 < -3(8) - 1
10 < -24 - 1
10 < -25 (false statement).
Therefore, (8, 10) is not a solution to the system.
