Answer:
The "a" tag or <a></a> and its "h<u>r</u>ef" attribute <a href="#"></a>
Explanation:
In html there is only one way to create a link to an external source or a bookmark. The tag is created using the element <a>.
The attribute "href" is then added to describe the link of where it is to point to.
Example:
<a href="google.com">Go to Google</a>
While the text in between the tag describes what the link is about.
Answer:
see explaination
Explanation:
#include<stdio.h>
/* Your solution goes here */
//Impllementation of SwapArrayEnds method
void SwapArrayEnds(int sortArray[],int SORT_ARR_SIZE){
//Declare tempVariable as integer type
int tempVariable;
if(SORT_ARR_SIZE > 1){
tempVariable = sortArray[0];
sortArray[0] = sortArray[SORT_ARR_SIZE-1];
sortArray[SORT_ARR_SIZE-1] = tempVariable;
}
}
int main(void) {
const int SORT_ARR_SIZE = 4;
int sortArray[SORT_ARR_SIZE];
int i = 0;
sortArray[0] = 10;
sortArray[1] = 20;
sortArray[2] = 30;
sortArray[3] = 40;
SwapArrayEnds(sortArray, SORT_ARR_SIZE);
for (i = 0; i < SORT_ARR_SIZE; ++i) {
printf("%d ", sortArray[i]);
}
printf("\n");
return 0;
}
Please go to attachment for the program screenshot and output
Answer:
The answer is "Option a"
Explanation:
A method is a technique, that associated with both a message and an object. It includes information, behavior, and actions of an interface defining, how the object can be used, and wrong choices can be described as follows:
- In option b, It includes parameters in the method.
- In option c, It contains a parameter, that may be one or more.
- In option d, It contains one parameter also.
1100110-101101 = 111001 = 57
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.
