(5 pt.) The name of a variable in the C programming language is a string that can contain uppercase letters, lowercase letters, digits, or underscores. Furthermore, the first character in the string must be a letter, either uppercase or lowercase, or an underscore. If the name of a variable is determined by its first 8 characters, how many different variables can be named in C

Answers

Answer 1
Answer:

Answer:

The different variable in C is 21213316700.

Explanation:

Given value:

Total value  = letters + underscore value

Total value  = 52 + 1

Total value =53

choice for first character = 53 letters +10 digits

first character = 63

choice for remaining characters

So,

Variable number With one 1 character = 53

Variable number With 2 character  = 53 × 63

Variable number With 3 character = 53 × 63²

Variable number With 4 character  = 53 × 63³

.

.

.

Variable number With 7 character  = 53 *  63^(7)

Total difference variable = 53 + 53 × 63+ 53 × 63²+ 53 × 63³+....+ 63^(7)

Total difference variable = 53(1 + 63 + 63²+ 63³+ .... + 63^(7))

Formula:

1+x+x^2+x^3+x^4+........x^n\n\n\Rightarrow  (x^((n+1)) -1)/(x-1)

Total difference variable

53 * (63^(7+1) -1 )/(63-1)\n\n53 * (63^(8) -1 )/(63-1) \n\n \therefore 63^8 = 2.48 * 10^(14)\n\n53 * (2.48 * 10^(14) -1  )/(62)\n\n53 * (1.48 * 10^(14))/(62)\n\n21213316700


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It is possible to force the Hardware Simulator to load a built-in version of chip Xxx. This can be done by:

Answers

Answer:

The answer is "It Specifies the built-in chip in the module header".

Explanation:

The hardware simulator is also an instrument about which the device physical models could be constructed, in the mainframe, which includes a console, a power supply, or multiple sockets for plugging regular modules.

  • It plays a specific role in the computing system, and it is also essential for each standard module.
  • In the hardware controller, it can be used to launch an integrated Xxx processor version, which can be achieved by listing the built-in chip throughout the module header.

The Clean Air Act Amendments of 1990 prohibit service-related releases of all ____________. A) GasB) OzoneC) MercuryD) Refrigerants

Answers

Answer:

D. Refrigerants

Explanation:

In the United States of America, the agency which was established by US Congress and saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, pesticide use, contamination, and hazardous waste spills is the Environmental Protection Agency (EPA). Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act .

The Clean Air Act Amendments of 1990 prohibit service-related releases of all refrigerants such as R-12 and R-134a. This ban became effective on the 1st of January, 1993.

Refrigerants refers to any chemical substance that undergoes a phase change (liquid and gas) so as to enable the cooling and freezing of materials. They are typically used in air conditioners, refrigerators, water dispensers, etc.

The Clean Air Act Amendments of 1990 prohibit service-related releases of all refrigerants. The correct option is D.

Thus, The Environmental Protection Agency (EPA), which was formed by US Congress, is the organization tasked with regulating all facets of pollution, environmental cleanup, pesticide use, contamination, and hazardous material spills.

All refrigerants, including R-12 and R-134a, are forbidden from being released during service under the Clean Air Act Amendments of 1990. On January 1st, 1993, this ban came into force.

Any chemical compound that transforms into a different phase (liquid or gas) to allow for the cooling or freezing of items is referred to as a refrigerant. They are frequently found in refrigerators, water dispensers, air conditioners, and other appliances.

Thus, The Clean Air Act Amendments of 1990 prohibit service-related releases of all refrigerants. The correct option is D.

Learn more about Clean air act, refer to the link:

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#SPJ3

A _____ area network is one step up from a _____ area network in geographical range.

Answers

Answer:

A metropolitan area network is one step up from a local area network in geographical range.

Explanation:

A metropolitan area network is one step up from a local area network in geographical range.

A network is a group of devices connected together for communication. Networks can be classified according to the area they cover, A metropolitan area network is smaller than a wide area network but larger than a local area network.

A local area network consist of computers network in a single place. A group of LAN network form a metropolitan network

A metropolitan network is a network across a city or small region. A group of  MAN form a wide area network.

/* ELEN 1301 Programming Assignment #5. Name : Your name. Student ID : Your student ID #. Due date : Due date Purpose of the program : Finding the average, minimum and maximum values of seven entered numbers. Use iomanip library to show 3 digits below decimal point. Section 1 : Enter the first number. Set min and max variables to the entered number. Section 2 : Enter the next number. If it is smaller than min, replace the min with the entered number. If it is bigger than max, replace the max with the entered number. Section 3 : Repeat section 2 five more times, so that you have seven numbers. Section 4 : Calculate the average of the seven numbers and show the result with 3 digits below decimal point. (2 points) Section 5 : Show the minimum number. (2 points) Section 6 : Show the maximum number. (2 points) */ #include #include using namespace std; int main() { int n1, n2, n3, n4, n5, n6, n7, min, max; double sum = 0; // Write the rest of the program. return 0; } // main

Answers

Answer:

Here is the C++ program:

#include<iostream> //to use input output functions

#include<iomanip> //to use setprecision

using namespace std; //to identify objects cin cout

int main() { //start of main function

     int n1, n2, n3, n4, n5, n6, n7, min, max;  // declare variables for 7 numbers, minimum value and maximum value

     double sum= 0; //declare variable to hold sum of 7 numbers

     double average;  //declare variable to hold average of 7 numbers

     cout<<"Enter first number: "; //prompts user to enter 1st number

     cin>>n1; //reads first number from user

     max = n1; //sets the first number to maximum

     min=n1; //sets the first number to minimum

     cout<<"Enter second number: "; //prompts user to enter 2nd number

     cin>>n2; //reads second number from user

     if(n2<min){ //if second number is less than min

          min=n2;      } //sets min to n2

     if(n2>max){ //if n2 is greater than max

          max = n2;      } //sets max to n2

     cout<<"Enter third number: ";  //prompts user to enter 3rd number

     cin>>n3; //reads third number from user

     if(n3<min){ //checks if n3 is greater than min

          min=n3;      } //sets n3 to min

     if(n3>max){ //checks if n3 is greater than max

          max = n3;      }      //sets max to n3

    cout<<"Enter fourth number: ";//prompts user to enter 4th number

     cin>>n4; //reads fourth number from user

     if(n4<min){  //if n4 is less than min

          min=n4;      }  //sets min to n4

     if(n4>max){  //if n4 is greater than max

          max = n4;      }  //sets max to n4

     cout<<"Enter fifth number: "; //prompts user to enter 5th number

     cin>>n5; //reads fifth number from user

     if(n5<min){  //if n5 is less than min

          min=n5;     }  //sets min to n5

     if(n5>max){  //if n5 is greater than max

          max = n5;      }  //sets max to n5

     cout<<"Enter sixth number: "; //prompts user to enter 6th number

     cin>>n6; //reads sixth number from user

     if(n6<min){ // if n6 is less than min

          min=n6;      }  //sets min to n6

     if(n6>max){  //if n6 is greater than max

          max = n6;      }  //sets max to n6

     cout<<"Enter seventh number: ";//prompts user to enter 7th number

     cin>>n7; //reads seventh number from user

     if(n7<min){  //if n7 is less than minimum number

          min=n7;      }  //assigns n7 to min

     if(n7>max){  //if n7 is greater than the maximum number

          max = n7;      }  //assigns n7 to max

     sum = n1+n2+n3+n4+n5+n6+n7;  //adds 7 numbers

     average = sum/7;  //computes average of 7 numbers

     cout<<"The average is: "<<fixed<<setprecision(3)<<average<<endl; //displays average value up to show 3 digits below decimal point using setprecision method of iomanip library

     cout<<"The maximum number is: "<<max<<endl; //displays maximum number of 7 numbers

     cout<<"The minimum number is: "<<min<<endl;  //displays miimum number of 7 numbers    

     return 0; }

Explanation:

The program is well explained in the comments attached to each statement of program. For example if

n1 = 3

n2 = 9

n3 = 7

n4 = 6

n5 = 2

n6 = 5

n7 = 4

When n1 is read using cin then this number is set to max and min as:

min = 9

max = 3

Now when n2 is read, the first if condition checks if n2 is less than min and second if condition checks if n2 is greater than max. As n2 = 9 so it is not less than min so this if condition is false and n2 is greater than max i.e. 3 so this condition is true. So 9 is assigned to max.

min = 9

max = 9

Now when n3 is read the values of min and max become:

min = 7

max = 9

Now when n4 is read the values of min and max become:

min = 6

max = 9

Now when n5 is read the values of min and max become:

min = 2

max = 9

Now when n6 is read the values of min and max become:

min = 2

max = 9

Now when n7 is read the values of min and max become:

min = 2

max = 9

Now the statement       sum = n1+n2+n3+n4+n5+n6+n7;

executes which becomes:

sum = 3 + 9 + 7 + 6 + 2 + 5 + 4

sum = 36.0

Next program control moves to statement:

average = sum/7;

this becomes

average = 36/7

5.142857

Since this is to be displayed up to 3 decimal places so average = 5.143

the complete output of the program is attached.

Consider the following two code segments, which are both intended to determine the longest of the three strings "pea", "pear", and "pearl" that occur in String str. For example, if str has the value "the pear in the bowl", the code segments should both print "pear" and if str has the value "the pea and the pearl", the code segments should both print "pearl". Assume that str contains at least one instance of "pea".I.

if (str.indexOf("pea") >= 0)

{

System.out.println("pea");

}

else if (str.indexOf("pear") >= 0)

{

System.out.println("pear");

}

else if (str.indexOf("pearl") >= 0)

{

System.out.println("pearl");

}

II.

if (str.indexOf("pearl") >= 0)

{

System.out.println("pearl");

}

else if (str.indexOf("pear") >= 0)

{

System.out.println("pear");

}

else if (str.indexOf("pea") >= 0)

{

System.out.println("pea");

}

Which of the following best describes the output produced by code segment I and code segment II?

Both code segment I and code segment II produce correct output for all values of str.

Neither code segment I nor code segment II produce correct output for all values of str.

Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pear" but not "pearl".

Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pearl".

Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".

Answers

Answer:

Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".

Explanation:

The main issue with the first code segment is the way how the if else if condition are arranged. The "pea" is checked at the earliest time in the first code segment and therefore so long as there is the "pea" exist in the string (regardless there is pear or pearl exist in the string as well), the if condition will become true and display "pea" to terminal. This is the reason why the code segment 1 only work for the values of str that contain "pea".

Write a loop that counts the number of space characters in a string. Recall that the space character is represented as ' '.

Answers

Answer:

I am writing Python program.

string = input("Enter a string: ")

print(string.count(' '))

Explanation:

The first statement takes input string from the user.

input() is used to read the input from the user.

The next statement uses count() function to count the number of times the specified object which is space ' ' hereoccurs in the string.

The print() function is used to return the number of times the space occurs in the string entered by the user.

Output:

Enter a string: How are you doing today?

4

The screenshot of program and its output is attached.