The problem with the media giving equal air time to those who are <u>not worthy</u> about the effects of media violence on violent behaviour is that the public is then misled into thinking that the evidence for such effects is <u>strong and unbiased</u> than it actually is.
<u>Explanation:</u>
Media has an extensive impact on the public. Whatever they air reaches each and every citizen of the specified region through direct and indirect means.
The media should take precautions and make sure whatever they are showing is unbiased and supported by true facts, figures, and objective evidence.
Answer / Explanation:
195.200.0.0/16
Note: Class C address can not be assigned a subnet mask of /16 because class c address has 24 bits assigned for network part.
2ⁿ = number of subnets
where n is additional bits borrowed from the host portion.
2ˣ - 2 = number of hosts
where x represent bits for the host portion.
Assuming we have 195.200.0.0/25
In the last octet, we have one bit for the network
number of subnets = 2¹ =2 network addresses
number of host = 2⁷ - 2= 126 network addresses per subnets
Answer:
cannot be used outside of its intented purpose
Explanation: python keywords are special reserved words that have specific meanings and purposes and can't be used for anything but those specific purposes
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:
False
Explanation:
JavaScript is language used along with html documents.
Java is a full-fledged programming language that handles applications.
