Look at the picture. I have no idea why I can't add an answer.
PLEASE HELP WILL GIVE BRANILYEST
1 answer:
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Answer:
The sample 2 has a lowest value of SE corresponding to the least sample variability.
Step-by-step explanation:
As the value of the sample means and standard deviations are not given, as similar question is found online from which the values of data is follows
The data is as attached with the solution. From this data
Sample 1 has a mean of 34 and a SE of 5
Sample 2 has a mean of 30 and a SE of 2
Sample 3 has a mean of 26 and a SE of 3
Sample 4 has a mean of 38 and a SE of 5
As per the measure of the sample variability is linked with the value of SE or standard error. Which is lowest in the case of sample 2 .
So the sample 2 has a lowest value of SE corresponding to the least sample variability.
Answer:
a. A(x) = (1/2)x(9 -x^2)
b. x > 0 . . . or . . . 0 < x < 3 (see below)
c. A(2) = 5
d. x = √3; A(√3) = 3√3
Step-by-step explanation:
a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...
A(x) = (1/2)(x)(y)
A(x) = (1/2)(x)(9-x^2)
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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).
On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).
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c. A(2) = (1/2)(2)(9 -2^2) = 5
The area is 5 when x=2.
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d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).
