Construct a frequency distribution and a frequency histogram for the data set using the indicated number of classes. Describe an
y patterns. Number of classes: 8 Data set: Reaction times (in milliseconds) of 30 adult females to an auditory stimulus
1 answer:
Answer:
Hello The data set is missing below is the missing data set
507,389,305,291,336,310,514,442,373,428,387,454,323,441,388,426,411,382,320,450,309,416,359,388,307,337,469,351,422,413
Step-by-step explanation:
Data set:
507,389,305,291,336,310,514,442,373,428,387,454,323,441,388,426,411,382,320,450,309,416,359,388,307,337,469,351,422,413
To CONSTRUCT a Frequency Distribution do the following
- Identify the Number of classes ( 8 ).
- Find the data range ( 514-291) = 223
- Find the width 223/8 = 27.9. ≈ 28
- create 8 groups each of 28 width starting from (291).
- count the no of data set in each interval and use that to generate the frequency table as attached below
attached is the frequency table and histogram representation
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Answer:
A. see below for a graph
B. f(x, y) = f(0, 15) = 90 is the maximum point
Step-by-step explanation:
A. See below for a graph. The vertices are those defined by the second inequality, since it is completely enclosed by the first inequality: (0, 0), (0, 15), (10, 0)
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B. For f(x, y) = 4x +6y, we have ...
f(0, 0) = 0
f(0, 15) = 6·15 = 90 . . . . . the maximum point
f(10, 0) = 4·10 = 40
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<em>Comment on evaluating the objective function</em>
I find it convenient to draw the line f(x, y) = 0 on the graph and then visually choose the vertex point that will put that line as far as possible from the origin. Here, the objective function is less steep than the feasible region boundary, so vertices toward the top of the graph will maximize the objective function.
