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pickupchik [31]
3 years ago
7

Normal or random variations that are considered part of operating the system at its current capability are ______: ​ a. systemic

variations. b. special cause variations. c. common cause variations. d. life cycle variations
Computers and Technology
1 answer:
Effectus [21]3 years ago
4 0

Normal or random variations that are considered part of operating the system at its current capability are <u> c. common cause variations.</u>

Explanation:

Common cause variation is fluctuation caused by unknown factors resulting in a steady but random distribution of output around the average of the data.

Common-cause variation is the natural or expected variation in a process.

Common-cause variation is characterised by:

  1. Phenomena constantly active within the system
  2. Variation predictable probabilistically
  3. Irregular variation within a historical experience base

It is a measure of the process potential, or how well the process can perform when special cause variation removed.

Common cause variation arises from external sources that are not inherent in the process and is where statistical quality control methods are most useful.

Statistical process control charts are used when trying to monitor and control 5- and 6-sigma quality levels.

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Answer:

<em>C++</em>

////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <iostream>

#include <vector>

using namespace std;

int main() {

  vector<int> v;

   int n = 1;

   while (n != 0) {

       cout<<"Enter an integer, the input ends if it is 0: ";

       cin>>n;

       v.push_back(n);

   }

   cout<<endl;

   ///////////////////////////////////////////////////////////

   int sum = 0;

   int num_positives = 0, num_negatives = 0;

   for (int i=0; i<v.size()-1; i++) {

       if (v[i] > 0)

           ++num_positives;

       else

           ++num_negatives;

           

       sum = sum + v[i];

   }

   //////////////////////////////////////////////////////////

   cout<<"The number of positives is "<<num_positives<<endl;

   cout<<"The number of negatives is "<<num_negatives<<endl;

   cout<<"The total is "<<sum<<endl;

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   ///////////////////////////////////////////////////////////

   return 0;

}

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Yakvenalex [24]
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Can anybody do Algorithm 2 for me (with Python).<br> Answer = 25 points.
Nostrana [21]

Answer:

age=int(input("Enter age"))

if age>=18:

    print("You are Young")

else

   print("You are child")

Explanation:

if you have any query or any problem kindly ask in comment

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frosja888 [35]

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Explanation:

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