Here's a C++ code snippet: #include <bits/stdc++.h> using namespace std; int main() { int n=50; int arr[n]; cout << "Enter 50 integers: "; for (int i = 0; i < n; i++) cin >> arr[i]; cout << endl; sort(arr,arr+n); for (int i = 0; i < n; i++) { cout << "pair which gives sum equal to " << arr[i] << " is:"; for (int j = 0; j < n; j++) for (int k = j + 1; k < n; k++) if (arr[i] == arr[j] + arr[k]) cout << "(" << arr[j] << " & " << arr[k] << ")" << ","; cout << endl; cout << "*******************************" << endl; } return 0; } The program will gather 50 integers from the user, sort them in ascending order, and identify all pairs that total to any of the array's elements, followed by printing those pairs.
Response:
a. Current disks do not reveal the actual locations of logical blocks.
Clarification:
Modern disks incorporate scheduling algorithms within the disk drive itself. This presents challenges for operating systems trying to optimize rotational latency. All scheduling methods end up having to perform similarly due to potential constraints faced by the operating system. Disks are typically accessed in physical blocks. Modern technology applies more electronic control to the disk.
Answer:
Refer to the explanation
Explanation:
Number of Mobiles=x1
Number of Workstations=x2
Constraints:
50x1 + 70x2 <= 1,60,000
250x1 + 700x2 <= 12,000,000
x1 >= 0 x2 >= 0
Objective function N(x1, x2) = 7800000000 x1 + 7200000000 x2
Excel formulae:
G17 = D9 * E7 + F9 * G7
G18= D10 * E7 * F10 * G7
G19= E7
G20= G7
G21= I17 * E7 + I18 * G7 (This will yield the Maximum)
And E7 and G7 will be the solution.
Set up your sheet as shown in the diagram and apply the formulas.
Access the tool and select solver. Upon opening the Solver dialog, confirm ‘Assume Linear model’ and ‘assume non-negative’. Click to solve the model and keep the solution.
Response: 1,500,000 bytes.
Clarification:
If we assume the image dimensions are 4000 pixels in width and 3000 pixels in height, the total uncompressed image will consist of 4000*3000= 12,000,000 pixels.
In the case of a binary image, each pixel can have only two values, which necessitates one bit for each pixel.
This indicates that we need to accommodate 12,000,000 bits.
Given that 1 byte equals 8 bits.
So, to store an uncompressed binary image sized 4000 x 3000 pixels, 12,000,000/8 bytes is required ⇒ 1,500,000 bytes.
