Answer:
Computer ethics is a part of practical philosophy concerned with how computing professionals should make decisions regarding professional and social conduct. Margaret Anne Pierce, a professor in the Department of Mathematics and Computers at Georgia Southern University has categorized the ethical decisions related to computer technology and usage into three primary influences:
The individual's own personal code.
Any informal code of ethical conduct that exists in the work place.
Exposure to formal codes of ethics.
Explanation:
Write an INSERT statement that adds this row to the Categories table:
CategoryName: Brass
Code the INSERT statement so SQL Server automatically generates the value for the CategoryID column.
Answer:
INSERT INTO categories (CategoryName)
VALUES ('Brass Code');
Explanation:
The SQL refers to the Structured Query Language in which the data is to be designed and maintained that occurred in the relational database management system i.e it is to be used for maintaining and query the database
Now the INSERT statement should be written as follows
INSERT INTO categories (CategoryName)
VALUES ('Brass Code');
What is the quick key to highlighting a column?
Ctrl + down arrow
Ctrl + Shift + down arrow
Right-click + down arrow
Ctrl + Windows + down arrow
The quick key to highlighting a column is the Ctrl + Shift + down arrow. Thus, option (b) is correct.
What is column?The term column refers to how data is organized vertically from top to bottom. Columns are groups of cells that are arranged vertically and run from top to bottom. A column is a group of cells in a table that are vertically aligned. The column is the used in the excel worksheet.
The quick key for highlighting a column is Ctrl + Shift + down arrow. To select downward, press Ctrl-Shift-Down Arrow. To pick anything, use Ctrl-Shift-Right Arrow, then Ctrl-Shift-Down Arrow. In the Move/Highlight Cells, the was employed. The majority of the time, the excel worksheet was used.
As a result, the quick key to highlighting a column is the Ctrl + Shift + down arrow. Therefore, option (b) is correct.
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Answer:
its (B) ctrl+shift+down arrow
hope this helps <3
Explanation:
The fact that the speed of a vehicle is lower than the prescribed limits shall relieve the driver from the duty to decrease speed when approaching and crossing an intersection. True or false
Explanation:
the answer is false ........
The fact that the speed of a vehicle is lower than the prescribed limits shall relieve the driver from the duty to decrease speed when approaching and crossing an intersection is not true.
What is intersection in the traffic rules?An intersection serves as the point where many lanes cross on the road which is a point where driver is required to follow the traffic rules.
However, the above statement is not true because , a driver can decide not decrease speed when approaching and crossing an intersection, even though it is a punishable offense in traffic rules.
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Consider the following Stack operations:
push(d), push(h), pop(), push(f), push(s), pop(), pop(), push(m).
Assume the stack is initially empty, what is the sequence of popped values, and what is the final state of the stack? (Identify which end is the top of the stack.)
Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
Cloud computing gives you the ability to expand and reduce resources according to your specific service requirement.
a. True
b. False
Answer:
a. True
Explanation:
Cloud computing can be defined as a type of computing that requires shared computing resources such as cloud storage (data storage), servers, computer power, and software over the internet rather than local servers and hard drives.
Generally, cloud computing offers individuals and businesses a fast, effective and efficient way of providing services.
In Computer science, one of the most essential characteristics or advantages of cloud computing is rapid elasticity.
By rapid elasticity, it simply means that cloud computing gives you the ability to expand and reduce resources according to your specific service requirement because resources such as servers can be used to execute a particular task and after completion, these resources can then be released or reduced.
Some of the examples of cloud computing are Google Slides, Google Drive, Dropbox, OneDrive etc.
Write a method named coinFlip that accepts as its parameter a string holding a file name, opens that file and reads its contents as a sequence of whitespace-separated tokens. Assume that the input file data represents results of sets of coin flips. A coin flip is either the letter H or T, or the word Heads or Tails, in either upper or lower case, separated by at least one space. You should read the sequence of coin flips and output to the console the number of heads and the percentage of heads in that line, rounded to the nearest whole number. If this percentage is 50% or greater, you should print a "You win!" message; otherwise, print "You lose!". For example, consider the following input file: H T H H T Tails taIlS tAILs TailS heads HEAds hEadS For the input above, your method should produce the following output: 6 heads (50%) You win!
Answer:
Here is the JAVA program:
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws FileNotFoundException{ //the start of main() function body, it throws an exception that indicates a failed attempt to open the file
Scanner input = new Scanner(new File("file.txt")); //creates a Scanner object and a File object to open and scan through the file.txt
coinFlip(input); } //calls coinFlip method
public static void coinFlip(Scanner input) { //coinFlip method that accepts as its parameter a string input holding a file name
while(input.hasNextLine()) { //iterates through the input file checking if there is another line in the input file
Scanner scan = new Scanner(input.nextLine()); //creates a Scanner object
int head = 0; // stores count of number of heads
int count = 0; //stores count of total number of tokens
while(scan.hasNext()) { //iterates through the sequence checking if there is another sequence in the input file
String token= scan.next(); // checks and returns the next token
if (token.equalsIgnoreCase("H")||token.equalsIgnoreCase("Heads")) { //compares H or Heads with the tokens in file ignoring lower case and upper case differences
head++; } //if a token i.e. any form of heads in file matches with the H or Heads then add 1 to the number of heads
count++; } //increment to 1 to compute total number of counts
double result = Percentage(head, count); //calls Percentage method passing number of heads and total counts to compute the percentage of heads
System.out.println(head + " heads " + "(" + result +"%)"); // prints the number of heads
if(result >= 50.00) { //if the percentage is greater or equal to 50
System.out.println("You win!");} //displays this message if above if condition is true
else //if the percentage is less than 50
{System.out.println("You lose!");} } } //displays this message if above if condition is false
public static double Percentage(int h, int total) { //method to compute the percentage of heads
double p = (double)h/total* 100; // divide number of heads with the total count and multiply the result by 100 to compute percentage
return p; } } //returns result
Explanation:
The program is well explained in the comments mentioned with each line of the above code. I will explain how the method coinFlip works.
Method coinFlip accepts a string holding a file name as its parameter. It opens that file and reads its contents as a sequence of tokens. Then it reads and scans through each token and the if condition statement:
if (token.equalsIgnoreCase("H")||token.equalsIgnoreCase("Heads"))
checks if the each token in the sequence stored in the file is equal to the H or Heads regardless of the case of the token. For example if the first token in the sequence is H then this if condition evaluates to true. Then the head++ statement increments the count of head by 1. After scanning each token in the sequence the variable count is also increased to 1.
If the token of the sequence is HeAds then this if condition evaluates to true because the lower or upper case difference is ignored due to equalsIgnoreCase method. Each time a head is found in the sequence the variable head is incremented to 1.
However if the token in the sequence is Tails then this if condition evaluates to false. Then the value of head variable is not incremented to 1. Next the count variable is incremented to 1 because this variable value is always incremented to 1 each time a token is scanned because count returns the total number of tokens and head returns total number of heads in the tokens.
Percentage method is used to return the percentage of the number of heads in the sequence. It takes head and count as parameters (h and total). Computes the percentage by this formula h/total* 100. If the result of this is greater than or equal to 50 then the message You win is displayed otherwise message You lose! is displayed in output.
A machine on a 10 Mbps network is regulated by a token bucket algorithm with a fill rate of 3 Mbps. The bucket is initially filled to capacity at 3MB. How long can the machine transmit at the full 10 Mbps capacity
Consider the following calling sequences and assuming that dynamic scoping is used, what variables are visible during execution of the last function called? Include with each visible variable the name of the function in which it was defined.a. Main calls fun1; fun1 calls fun2; fun2 calls fun3b. Main calls fun1; fun1 calls fun3c. Main calls fun2; fun2 calls fun3; fun3 calls fun1d. Main calls fun3; fun3 calls fun1e. Main calls fun1; fun1 calls fun3; fun3 calls fun2f. Main calls fun3; fun3 calls fun2; fun2 calls fun1void fun1(void);void fun2(void);void fun3(void);void main() {Int a,b,c;…}void fun1(void){Int b,c,d;…}void fun2(void){Int c,d,e;…}void fun3(void){Int d,e,f;…}
Answer:
In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.
Explanation:
In main() function three integer type variables are declared: a,b,c
In fun1() three int type variables are declared/defined: b,c,d
In fun2() three int type variables are declared/defined: c,d,e
In fun3() three int type variables are declared/defined: d,e,f
a. Main calls fun1; fun1 calls fun2; fun2 calls fun3
Here the main() calls fun1() which calls fun2() and fun2() calls func3() . This means first the func3() executes, then fun2(), then fun1() and last main()
Visible Variable: d, e, f Defined in: fun3
Visible Variable: c Defined in: fun2 (the variables d and e of fun2
are not visible)
Visible Variable: b Defined in: fun1 ( c and d of func1 are hidden)
Visible Variable: a Defined in: main (b,c are hidden)
b. Main calls fun1; fun1 calls fun3
Here the main() calls fun1, fun1 calls fun3. This means the body of fun3 executes first, then of fun1 and then in last, of main()
Visible Variable: d, e, f Defined in: fun3
Visible Variable: b, c Defined in: fun1 (d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
c. Main calls fun2; fun2 calls fun3; fun3 calls fun1
Here the main() calls fun2, fun2 calls fun3 and fun3 calls fun1. This means the body of fun1 executes first, then of fun3, then fun2 and in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e, f Defined in: fun3 ( d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
Here variables c, d and e of fun2 are not visible
d. Main calls fun3; fun3 calls fun1
Here the main() calls fun3, fun3 calls fun1. This means the body of fun1 executes first, then of fun3 and then in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e, f Defined in: fun3 ( d not visible )
Visible Variable: a Defined in: main (b and c not visible)
e. Main calls fun1; fun1 calls fun3; fun3 calls fun2
Here the main() calls fun1, fun1 calls fun3 and fun3 calls fun2. This means the body of fun2 executes first, then of fun3, then of fun1 and then in last, of main()
Visible Variable: c, d, e Defined in: fun2
Visible Variable: f Defined in: fun3 ( d and e not visible)
Visible Variable: b Defined in: fun1 ( c and d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
f. Main calls fun3; fun3 calls fun2; fun2 calls fun1
Here the main() calls fun3, fun3 calls fun2 and fun2 calls fun1. This means the body of fun1 executes first, then of fun2, then of fun3 and then in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e Defined in: fun2
Visible Variable: f Defined in: fun3
Visible Variable: a Defined in: main
CHALLENGE ACTIVITY 3.7.2: Type casting: Reading and adding values.
Assign totalowls with the sum of num_owls A and num_owls_B.
Sample output with inputs: 34
Number of owls: 7
1. total_owls -
2.
3. num_owls A - input
4. num_owls_B - input
5.
6. " Your solution goes here
7.
8. print("Number of owls:', total_owls)
Answer:
total_owls = 0
num_owls_A = int(input())
num_owls_B = int(input())
total_owls = num_owls_A + num_owls_B
print("Number of owls:", total_owls)
Explanation:
Initialize the total_owls as 0
Ask the user to enter num_owls_A and num_owls_B, convert the input to int
Sum the num_owls_A and num_owls_B and set it to the total_owls
Print the total_owls
avg_owls=0.0
num_owls_zooA = int(input())
num_owls_zooB = int(input())
num_owls_zooC = int(input())
num_zoos = 3
avg_owls = (num_owls_zooA + num_owls_zooB + num_owls_zooC) / num_zoos
print('Average owls per zoo:', int(avg_owls))
There is a problem while you are trying the calculate the average owls per zoo.
As you may know, in order to calculate the average, you need to get the total number of owls in all the zoos, then divide it to the number of zoos. That is why you need to sum num_owls_zooA + num_owls_zooB + num_owls_zooC and divide the result by num_zoos
Be aware that the average is printed as integer value. That is why you need to cast the avg_owls to an int as int(avg_owls)
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