Answer:
(D) Expression I and expression Il evaluate to the same value only when A and B differ.
Explanation:
Given
[tex]A\ \&\&\ B[/tex]
[tex]!A\ \&\&\ !B[/tex]
Required
Select the true statement
To do this, I will create the following case scenarios.
(a): [tex]A = true[/tex] and [tex]B = true[/tex]
[tex]A\ \&\&\ B[/tex]
[tex]true\ \&\&\ true \to true[/tex] i.e. true and true is true
[tex]!A\ \&\&\ !B[/tex]
[tex]!true\ \&\&\ !true[/tex]
------------------------------------------------------
[tex]!true = false[/tex]
------------------------------------------------------
So, we have:
[tex]false\ \&\&\ false \to false[/tex]
So:
[tex]A =true[/tex] and [tex]B = true[/tex]
[tex]A\ \&\&\ B = true[/tex]
[tex]!A\ \&\&\ !B = false[/tex]
Hence, options (B) and (E) are incorrect
(b): [tex]A = true[/tex] and [tex]B = false[/tex]
[tex]A\ \&\&\ B[/tex]
[tex]true\ \&\&\ false \to false[/tex]
[tex]!A\ \&\&\ !B[/tex]
[tex]!true\ \&\&\ !false[/tex]
Solve each negation
[tex]false\ \&\&\ true \to false[/tex]
So:
[tex]A =true[/tex] and [tex]B = false[/tex]
[tex]A\ \&\&\ B = false[/tex]
[tex]!A\ \&\&\ !B = false[/tex]
Hence, option (c) is incorrect
(c): [tex]A = false[/tex] and [tex]B = true[/tex]
[tex]A\ \&\&\ B[/tex]
[tex]false\ \&\&\ true \to false[/tex]
[tex]!A\ \&\&\ !B[/tex]
[tex]!false\ \&\&\ !true[/tex]
[tex]truee\ \&\&\ false \to false[/tex]
So:
[tex]A =false[/tex] and [tex]B = true[/tex]
[tex]A\ \&\&\ B = false[/tex]
[tex]!A\ \&\&\ !B = false[/tex]
Case scenarios b and c implies that option (d) is correct because different values of A and B gives the same value of both expression which is false
This also implies that (a) is incorrect.
How I did it. Part 4
Answer:
What are you asking for in this problem?
Explanation:
c Write a program that simulates a magic square using 3 one dimensional parallel arrays of integer type. Each one the arrays corresponds to a row of the magic square. The program asks the user to enter the values of the magic square row by row and informs the user if the grid is a magic square or not. flowchart
Answer:
hope this helps and do consider giving a brainliest to the ans if it helped.
Explanation:
//program to check if the entered grid is magic square or not
/**c++ standard libraries
*/
#include<bits/stdc++.h>
using namespace std;
/**function to check whether the entered grid is magic square or not
*/
int isMagicSquare(int arr[3][3]){
int i,j,sum=0,sum1=0,rsum,csum;
for(i=0;i<3;i++){
sum+=arr[i][i];
sum1+=arr[i][2-i];
}
if(sum!=sum1){
return 0;
}
for(i=0;i<3;i++){
rsum=0;
csum=0;
for(j=0;j<3;j++){
rsum+=arr[i][j];
csum+=arr[j][i];
}
if(sum!=rsum){
return 0;
}
if(sum!=csum){
return 0;
}
}
return 1;
}
/** main function to get user entries and
* call function
* and print output
*/
int main(){
int i,j,arr[3][3]={0};
for(i=0;i<3;i++){
for(j=0;j<3;j++){
cout<<"Enter the number for row "<<i<<" and column "<<j<<" : ";
cin>>arr[i][j];
}
}
int ret = isMagicSquare(arr);
if(ret==1){
cout<<"This is a Lo Shu magic square"<<endl;
}
else{
cout<<"This is not a Lo Shu magic square"<<endl;
}
return 0;
}
Write a Java program named Problem 3 that prompts the user to enter two integers, a start value and end value ( you may assume that the start value is less than the end value). As output, the program is to display the odd values from the start value to the end value. For example, if the user enters 2 and 14, the output would be 3, 5, 7, 9, 11, 13 and if the user enters 14 and 3, the output would be 3, 5, 7, 9, 11, 13.
Answer:
hope this helps
Explanation:
import java.util.Scanner;
public class Problem3 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Enter start value: ");
int start = in.nextInt();
System.out.print("Enter end value: ");
int end = in.nextInt();
if (start > end) {
int temp = start;
start = end;
end = temp;
}
for (int i = start; i <= end; i++) {
if (i % 2 == 1) {
System.out.print(i);
if (i == end || i + 1 == end) {
System.out.println();
} else {
System.out.print(", ");
}
}
}
}
}
Which of the following is NOT an example of a metasearch engine?
Answer:
Where is the choices mate?
What is the problem with paying only your minimum credit card balance each month?
A. It lowers your credit score
B. You have to pay interest
C. The bank will cancel your credit card
D. All of the above
Answer:b
Explanation:
Answer: The answer is B.
Explanation: While yes it is important to make at least the minimum payment, its not ideal to carry a balance from month to month (You'll rack up interest charges). And risk falling into debt. Therefore, the answer is B.
Hope this helps!
In numerical methods, one source of error occurs when we use an approximation for a mathematical expression that would otherwise be too costly to compute in terms of run-time or memory resources. One routine example is the approximation of infinite series by a finite series that mostly captures the important behavior of the infinite series.
In this problem you will implement an approximation to the exp(x) as represented by the following infinite series,
exp(x) = Infinity n= 0 xn/xn!
Your approximation will be a truncated finite series with N + 1 terms,
exp(x, N) = Infinityn n = 0 xn/xn!
For the first part of this problem, you are given a random real number x and will investigate how well a finite series expansion for exp(x) approximates the infinite series.
Compute exp(x) using a finite series approximation with N E [0, 9] N (i.e. is an integer).
Save the 10 floating point values from your approximation in a numpy array named exp_approx. exp_approx should be of shape (10,) and should be ordered with increasing N (i.e. the first entry of exp_approx should correspond to exp(x, N) when N = 0 and the last entry when N = 9).
Answer:
The function in Python is as follows:
import math
import numpy as np
def exp(x):
mylist = []
for n in range(10):
num = (x**n)/(math.factorial(n))
mylist.append([num])
exp_approx = np.asarray(mylist)
sum = 0
for num in exp_approx:
sum+=num
return sum
Explanation:
The imports the python math module
import math
The imports the python numpy module
import numpy as np
The function begins here
def exp(x):
This creates an empty list
mylist = []
This iterates from 0 to 9
for n in range(10):
This calculates each term of the series
num = (x**n)/(math.factorial(n))
This appends the term to list, mylist
mylist.append([num])
This appends all elements of mylist to numpy array, exp_approx
exp_approx = np.asarray(mylist)
This initializes the sum of the series to 0
sum = 0
This iterates through exp_approx
for num in exp_approx:
This adds all terms of the series
sum+=num
This returns the calculated sum
return sum
In c please
Counting the character occurrences in a file
For this task you are asked to write a program that will open a file called “story.txt
and count the number of occurrences of each letter from the alphabet in this file.
At the end your program will output the following report:
Number of occurrences for the alphabets:
a was used-times.
b was used - times.
c was used - times .... ...and so, on
Assume the file contains only lower-case letters and for simplicity just a single
paragraph. Your program should keep a counter associated with each letter of the
alphabet (26 counters) [Hint: Use array|
| Your program should also print a histogram of characters count by adding
a new function print Histogram (int counters []). This function receives the
counters from the previous task and instead of printing the number of times each
character was used, prints a histogram of the counters. An example histogram for
three letters is shown below) [Hint: Use the extended asci character 254]:
Answer:
#include <stdio.h>
#include <ctype.h>
void printHistogram(int counters[]) {
int largest = 0;
int row,i;
for (i = 0; i < 26; i++) {
if (counters[i] > largest) {
largest = counters[i];
}
}
for (row = largest; row > 0; row--) {
for (i = 0; i < 26; i++) {
if (counters[i] >= row) {
putchar(254);
}
else {
putchar(32);
}
putchar(32);
}
putchar('\n');
}
for (i = 0; i < 26; i++) {
putchar('a' + i);
putchar(32);
}
}
int main() {
int counters[26] = { 0 };
int i;
char c;
FILE* f;
fopen_s(&f, "story.txt", "r");
while (!feof(f)) {
c = tolower(fgetc(f));
if (c >= 'a' && c <= 'z') {
counters[c-'a']++;
}
}
for (i = 0; i < 26; i++) {
printf("%c was used %d times.\n", 'a'+i, counters[i]);
}
printf("\nHere is a histogram:\n");
printHistogram(counters);
}