07.06.2021

Information or signals entered into a computer system is

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24.06.2023, solved by verified expert
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Logs.

Like data logs. Sometimes people make these logs to keep tabs on other people or to get important information put down somewhere that way it is saved and can be looked back upon later. Anytime someone makes an action on the computer, it makes a TMP file representing a log of what you want it to do before the computer quickly get's rid of the file.

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Engineering
Step-by-step answer
P Answered by PhD

Logs.

Like data logs. Sometimes people make these logs to keep tabs on other people or to get important information put down somewhere that way it is saved and can be looked back upon later. Anytime someone makes an action on the computer, it makes a TMP file representing a log of what you want it to do before the computer quickly get's rid of the file.

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Mathematics
Step-by-step answer
P Answered by Specialist

A) Most points are between these limits, so the process appears to be in control with respect to variability.

D) The value of S on the 5th day lies above the UCL, so an out-of-control signal is generated.

Step-by-step explanation:

We have the data of 30 days of daily samples, of size n=200. The data is expressed in "number of non-conforming chips". To graph the data in a p-chart, we have to calculate the proportion, so we have to divide the number of non-conforming chips in each sample by the sample size in order to get the proportion.

The data of non-conforming chips es:

[916241836196268253017161915201322]

The proportion of non-conforming chips is calculated as:

p_i=\dfrac{x_i}{n}=\dfrac{x_i}{200}

Then, the proportions of non-conforming chips are:

[0.0450.080.120.090.180.0950.030.130.040.1250.150.0850.080.0950.0750.10.0650.110.0450.1150.090.0750.0750.1250.160.0950.060.1150.0750.13

]

For the p-chart we calculate the average proportion:

\bar{p}=\dfrac{1}{30}\sum p_i=\dfrac{2.855}{30}\approx0.095

Then, we calculate the control limits as:

\bar{p}\pm\sqrt{\dfrac{\bar{p}(1-\bar{p})}{m}}\\\\\\0.095\pm\sqrt{\dfrac{0.095\cdot0.905}{30}}=0.095\pm0.054\\\\\\LCL=0.095-0.054=0.041\\\\UCL=0.095+0.054=0.149

With these values, we can graph the data in the p-chart.

We can see that most points are within these limits, as it is expected (we are calculating the control limits with the same data we are charting). There are 5 out of 30 (one sixth of the samples) that are outside the control limits but there are not more than one consecutive points outside the contorl limits, so we can conclude that the process is in control.

The value of the 5th day (p=0.180) is over the UCL (UCL=0.149), so an out-of-control signal is generated. As the 6th day sample is in control (p=0.095), we could conclude that a special condition makes the 5th day sample lies outside the control limits.


A sample of 200 ROM computer chips was selected on each of 30 consecutive days, and the number of no
Mathematics
Step-by-step answer
P Answered by Master

A) Most points are between these limits, so the process appears to be in control with respect to variability.

D) The value of S on the 5th day lies above the UCL, so an out-of-control signal is generated.

Step-by-step explanation:

We have the data of 30 days of daily samples, of size n=200. The data is expressed in "number of non-conforming chips". To graph the data in a p-chart, we have to calculate the proportion, so we have to divide the number of non-conforming chips in each sample by the sample size in order to get the proportion.

The data of non-conforming chips es:

[916241836196268253017161915201322]

The proportion of non-conforming chips is calculated as:

p_i=\dfrac{x_i}{n}=\dfrac{x_i}{200}

Then, the proportions of non-conforming chips are:

[0.0450.080.120.090.180.0950.030.130.040.1250.150.0850.080.0950.0750.10.0650.110.0450.1150.090.0750.0750.1250.160.0950.060.1150.0750.13

]

For the p-chart we calculate the average proportion:

\bar{p}=\dfrac{1}{30}\sum p_i=\dfrac{2.855}{30}\approx0.095

Then, we calculate the control limits as:

\bar{p}\pm\sqrt{\dfrac{\bar{p}(1-\bar{p})}{m}}\\\\\\0.095\pm\sqrt{\dfrac{0.095\cdot0.905}{30}}=0.095\pm0.054\\\\\\LCL=0.095-0.054=0.041\\\\UCL=0.095+0.054=0.149

With these values, we can graph the data in the p-chart.

We can see that most points are within these limits, as it is expected (we are calculating the control limits with the same data we are charting). There are 5 out of 30 (one sixth of the samples) that are outside the control limits but there are not more than one consecutive points outside the contorl limits, so we can conclude that the process is in control.

The value of the 5th day (p=0.180) is over the UCL (UCL=0.149), so an out-of-control signal is generated. As the 6th day sample is in control (p=0.095), we could conclude that a special condition makes the 5th day sample lies outside the control limits.


A sample of 200 ROM computer chips was selected on each of 30 consecutive days, and the number of no
Engineering
Step-by-step answer
P Answered by Master

True, a retractable service pit cover can help other shop employees from falling into the pit while another technician is in the pit servicing a car.

Retractable service pit cover is a special cover made to cover service pit which aid inspection and repair of vehicle in an auto workshop.

Retractable service pit ensure safety of workshop technicians around service pits.Retractable service pits aid proper inspection of the beneath of the vehicleRetractable service pits enable the technician to have a close view of the vehicle in repair.The pit also serves as guide to short-eyed people or people to are not aware of the structure.In essence, the pit prevent against injury or death of workers and occupier.

In conclusion, the retractable service pit cover help other shop employees from falling into the pit even if they are aware of the structure.

Learn more about retractable service pit in picture attached

Engineering
Step-by-step answer
P Answered by PhD

Code:

#include <iostream>

using namespace std;

int main()

{

  int Car_Year;

  cout<<"Please Enter the Car Model."<<endl;

  cin>>Car_Year;    

 if (Car_Year<1967)

 {

cout<<"Few safety features."<<endl;

 }

else if (Car_Year>1971 && Car_Year<=1991)

{

cout<<"Probably has head rests."<<endl;

}

else if (Car_Year>1991 && Car_Year<=2000)

{

cout<<"Probably has antilock brakes."<<endl;

}

else if (Car_Year>2000)

{

cout<<"Probably has airbags."<<endl;

  }

else

{

cout<<"Invalid Selection."<<endl;

}

  return 0;

}

Output:

Please Enter the Car Model.

1975

Probably has head rests.

Please Enter the Car Model.

1999

Probably has antilock brakes.

Please Enter the Car Model.

2005

Probably has airbags.

Please Enter the Car Model.

1955

Few safety features.

Explanation:

We were required to implement multiple If else conditions to assign car model year with the corresponding features of the car.

The code is tested with a wide range of inputs and it returned the same results as it was asked in the question.

Engineering
Step-by-step answer
P Answered by Master

"Test Phase " is the correct choice.

Explanation:

DevSecOps seems to be a community as well as experience of corporate data science which encompasses software design, regulation, including operational activities. This same main feature of DevSecOps has always been to strengthen customer achievement as well as expedition importance by computerizing, supervising as well as implementing data protection at all stages of the development including its development tools.The testing method throughout the test phase would then help make sure that the controller is designed mostly under the responsibilities forecasted. The test focuses on either the reaction times, dependability, use of resources but instead interoperability of applications.
Engineering
Step-by-step answer
P Answered by PhD

The following code or the program will be used

Explanation:

def readFile(filename):

   dict = {}

   with open(filename, 'r') as infile:

       lines = infile.readlines()

       for index in range(0, len(lines) - 1, 2):

           if lines[index].strip()=='':continue

           count = int(lines[index].strip())

           name = lines[index + 1].strip()

           if count in dict.keys():

               name_list = dict.get(count)

               name_list.append(name)

               name_list.sort()

           else:

               dict[count] = [name]

           print(count,name)

   return dict

def output_keys(dict, filename):

   with open(filename,'w+') as outfile:

       for key in sorted(dict.keys()):

           outfile.write('{}: {}\n'.format(key,';'.join(dict.get(key

           print('{}: {}\n'.format(key,';'.join(dict.get(key  

def output_titles(dict, filename):

   titles = []

   for title in dict.values():

       titles.extend(title)

   with open(filename,'w+') as outfile:

       for title in sorted(titles):

           outfile.write('{}\n'.format(title))

           print(title)

def main():

   filename = input('Enter input file name: ')

   dict = readFile(filename)

   if dict is None:

       print('Error: Invalid file name provided: {}'.format(filename))

       return

   print(dict)

   output_filename_1 ='output_keys.txt'

   output_filename_2 ='output_titles.txt'

   output_keys(dict,output_filename_1)

   output_titles(dict,output_filename_2)  

main()

Engineering
Step-by-step answer
P Answered by PhD

Energy equation from this week’s notes, your answer from #5, and Plank’s constant (6.63E-34) to find the approximate energy of this photon

Explanation:

1.The amount of energy in those photons is calculated by this equation, E = hf, where E is the energy of the photon in Joules; h is Planck's constant, which is always 6.63 * 10^-34 Joule seconds; and f is the frequency of the light in hertz

2.The first is Planck's equation, which was proposed by Max Planck to describe how energy is transferred in quanta or packets. Planck's equation makes it possible to understand blackbody radiation and the photoelectric effect. The equation is:

E = hν

where

E = energy

h = Planck's constant = 6.626 x 10-34 J·s

ν = frequency

Engineering
Step-by-step answer
P Answered by PhD

x=float(input("Enter a number: "))

sub=(x-int(x))

print(sub)

Explanation:

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