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ExtremeBDS [4]
4 years ago
13

Consider the simplicity of these two DOE examples. Next, consider the assertion that DOE isn’t put into practice as often or dee

ply as experts feel it should be. Describe and discuss the roles that simple examples like these might play – both positive and negative – in contributing to this lack of acceptance and penetration of the DOE techniques in practice. For example, is there obviousness to the conclusions of these simple examples that might actually contribute to hurting people’s perceptions of DOE? In providing such simple examples, do we sometimes actually inhibit management’s ability to properly visualize the true value that DOE could provide in more realistic (and therefore complicated) situations? Do these types of examples help support the required early learning, and, if so, are there typically articles out there that support learning at the next level of complexity? Try to draw a conclusion about why DOE isn’t practiced more widely in most workplaces.
Engineering
1 answer:
Bingel [31]4 years ago
5 0

Answer:

Do Re Mi

Explanation:

I have no clue sorry

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A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0
3 years ago
Air initially at 15 psla and 60 F is compressed to 75 psia and 400 F. The power input to air under steady state condition is 5 h
Triss [41]

Answer:\dot{m}=3.46lbm/min

Explanation:

Initial conditions

P_1=15 psia

T_1=60 F^{\circ}

Final conditions

P_2=75 psia

T_2=400F^{\circ}

Steady flow energy equation

\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}

\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}

\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121

-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ]

-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec

\dot{m}=0.057821lbm/sec

\dot{m}=3.46lbm/min

3 0
3 years ago
Scheduling can best be defined as the process used to determine:​
shutvik [7]

Answer:

Overall project duration

Explanation:

Scheduling can best be defined as the process used to determine a overall project duration.

8 0
3 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
erica [24]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

6 0
3 years ago
The following are related to wastewater treatment:
Paraphin [41]

Answer:

A. - Primary wastewater treatment: the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater.

- Secondary wastewater treatment: treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process.

-Tertiary wastewater treatment: Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending on the type of water contamination that must be removed.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C. - Activated Sludge, Rotating biological contractors.

Explanation:

A. Primary wastewater treatment refers to sedimentation, the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater. The main goal of primary treatment is to remove organic and inorganic solids that settles by sedimentation, and by act of skimming removed materials end up floating.

Secondary wastewater treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process. The goal of secondary treatment is the further treatment of the effluent from primary treatment in order to get rid of the residual organics and suspended solids.

Advanced wastewater is quite in between. Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending upon the type of water contamination that must be removed. Advanced treatment processes can be combined with primary or secondary treatment or used instead of secondary treatment.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C.

-Activated Sludge: The activated sludge process is a kind of wastewater treatment process for treating sewage/industrial wastewaters by the use of aeration and a biological floc which consists of bacteria and protozoa.

- Rotating biological contactor: Rotating biological contactors (RBCs) are fixed-film reactors that have similarities with biofilters in the sense that organisms are attached to support media.

Activated sludge is a suspended process, which is when the biomass is mixed with the sewage while in rotating biological contractor, the biomass grows on the media and then, the sewage passes over the surface.

6 0
4 years ago
Read 2 more answers
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