4.5 code practice computer science

What is the correct order of precedence of the mathematical operators?

LenaWriter [7]

A common technique for remembering the order<span> of </span>operations<span> is the abbreviation "PEMDAS", which is turned into the phrase "Please Excuse My Dear Aunt Sally". It stands for "Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction".</span>

7 0

3 years ago

By applying styles,____ formats are being applied each time?

Ray Of Light [21]

(A) the same, by applying styles the same formats are being applied each time.

8 0

3 years ago

Which of the following statements correctly describe a computer program?

Naddik [55]

Answer: A. It’s a robust way to find information

Explanation:

Hope this help.

3 0

3 years ago

Read 2 more answers

Define a method named swapValues that takes an array of four integers as a parameter, swaps array elements at indices 0 and 1, a

Luba_88 [7]

The program is an illustration of arrays.

Arrays are used to hold multiple values.

The program in java, where comments are used to explain each line is as follows:

import java.util.*;

public class Main{

//This defines the method

public static int[] swapValues(int[] arr) {

//This swaps the first and second array elements

int temp = arr[0];

arr[0] = arr[1]; arr[1] = temp;

//This swaps the third and fourth array elements

temp = arr[2];

arr[2] = arr[3]; arr[3] = temp;

//This returns the swapped array to main

return arr;

}

//The main method begins here

public static void main(String[] args) {

//This creates a Scanner object

Scanner input = new Scanner(System.in);

//This declares an array of 4 elements

int[] intArray = new int[4];

//This gets input for the array

for(int i = 0; i<4;i++){

intArray[i] = input.nextInt();

}

//This calls the swapValues method

intArray=swapValues(intArray);

//This prints the swapped array

for (int i = 0; i < 4; i++){

System.out.print( intArray[i]+ " "); }

}

At the end of the program, the elements are swapped and printed.

Read more about similar programs at:

brainly.com/question/14017034

6 0

3 years ago

We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is

bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons to find the middle terms.

If list has 18 elements, there are two middle terms as one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons to find the middle terms.

If list has 20 elements, there are two middle terms as one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

6 0

3 years ago