Information or signals entered into a computer system is

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.

-Ɽ3₮Ɽ0 Ⱬ3Ɽ0

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:

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:

Then, we calculate the control limits as:

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) 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:

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:

Then, we calculate the control limits as:

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.

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.

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

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.

"Test Phase " is the correct choice.

Explanation:

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

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()

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