Answer:
D. Zero-day
Explanation:
It is clearly stated that the antivirus is still up-to-date on the workstation, and there has been no alterations to the firewall. Hence, this means the software is functional and up-to-date with all known viruses. This shows that this attack is unknown to the manufacturer of the antivirus software and hence no virus definition or patch fixing this problem has been developed yet. In this light, it is a zero-day.
A zero-day is a type of vulnerability in a software that is not yet known to the vendor or manufacturer, hence this security hole makes the software vulnerable to attackers or hacker before it is been fixed.
Using the knowledge of pseudocodes it will be possible to write a code that calculates the amount of hours worked and giving warnings about it.
<h3>Writing a pseudocode we have that:</h3>
<em>while </em>
<em>if number == 0</em>
<em>break</em>
<em>hours =0 </em>
<em>for i =1 to 5 </em>
<em>hours = hours + time_out[ i ] - time_in[ i ]</em>
<em> If hours >36 : </em>
<em>print ( "Name is " ,name)</em>
<em>print( "No of hours are ",hours)</em>
<em>print("Congratulaion! Your working hours are more than 36")</em>
<em>If hours <30 : #</em>
<em>print ( "Name is " ,name)</em>
<em>print( "No of hours are ",hours)</em>
<em>print("Warning !!!")</em>
<em>End loop</em>
See more about pseudocode at brainly.com/question/13208346
#SPJ1
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
A-because most of the victims of intellectual theft are individuals
Explanation:
Its definitely the first, and not B., C, or D,
Countries are no barrier in catching the theft, as almost all the countries now have issued the online identification number of each of their citizens. However, some of them are in the process. Hence, for next one - two years we can add add the B as well:
B-because much intellectual theft is committed in other countries
However, as this work is complete, the B option will not be an issue related to intellectual theft analysis and control.
And C and D are definitely not the part of the answer, as associations like FCC USA, Internet association etc are technically sound enough, and quite good to fight a case under any law as well.
