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enyata [817]
3 years ago
12

ANOVA is a hypothesis testing technique used to compare the equality of means for two or more groups; for example, it can be use

d to test that the mean number of computer chips produced by a company on each of the day, evening, and night shifts is the same. Give an example of an application of ANOVA in an industrial setting that is different from the examples provided in the course. Discuss and share this information with your classmates.
Mathematics
1 answer:
Elza [17]3 years ago
6 0

Answer:

The answer is below

Step-by-step explanation:

ANOVA can be used in a construction industry whereby the construction or project manager is saddled with the responsibility of picking the most appropriate method of construction or materials to be used during construction in the face of two or more alternative options.

For example in terms of method of construction, two or three methods of construction may be available to pick from to construct a building project.

However, due to the possibility of differences in construction cost, and project duration. The construction manager may need to use an ANOVA test to determine the project duration of each method and if it is later revealed through the test that all the methods of construction will lead to the project being delivered on time as the client scheduled.

Then, the construction manager may then result to pick the least expensive alternative of the construction methods.

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A total of 937 people attended the play.
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6 0
3 years ago
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

5 0
3 years ago
NEED HELP ASAP ! PLEASE
solong [7]

Answer:

C'D'=10

Step-by-step explanation:

CB=5, and C'B'=12.5

5*2.5=12.5

4*2.5=10

8 0
2 years ago
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