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:

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.

The quick key to highlighting a column is the Ctrl + Shift + down arrow. Thus, option (b) is correct.

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.

Learn more about the column, here:

https://brainly.com/question/3642260

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Answer:

its (B) ctrl+shift+down arrow

hope this helps <3

Explanation:

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

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.

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