Bertha should use end roll technique so that she doesn't miss the initial action of a drama.
b. end roll
<u>Explanation:</u>
End roll is a simple method to decide whether the film is pushing ahead or not. When you utilize the film advance to wind the film, you essentially need to check if the handle on the left (that you use to rewind the film) is turning.
In the event that it turns, great, it implies that the film is appropriately locked in. So Bertha should utilize the end move strategy with the goal that she doesn't miss the underlying activity of a dramatization.
In this program, I am using the school-based grading system and the program should accept the subject and the number of students.
Program approach:-
- Using the necessary header file.
- Using the standard I/O namespace function.
- Define the main function.
- Declare the variable.
- Display enter obtain marks in 5 subjects.
- Return the value.
Program:-
//header file
#include<iostream>
//using namespace
using namespace std;
//main method
int main()
{
//declare variable
int j;
float mark, sum=0, a;
//display enter obtain marks in 5 subjects
cout<<"Enter Marks obtained in 5 Subjects: ";
for(j=0; j<5; j++)
{
cin>>mark;
sum = sum+mark;
}
a = sum/5;
//display grade
cout<<"\nGrade = ";
if(a>=91 && a<=100)
//display a1
cout<<"a1";
else if(a>=81 && a<91)
//display a2
cout<<"a2";
else if(a>=71 && a<81)
cout<<"b1";
else if(a>=61 && a<71)
cout<<"b2";
else if(a>=51 && a<61)
//display c1
cout<<"c1";
else if(a>=41 && a<51)
//display c2
cout<<"c2";
else if(a>=33 && a<41)
//display d
cout<<"d";
else if(a>=21 && a<33)
//display e1
cout<<"e1";
else if(a>=0 && a<21)
//display e2
cout<<"e2";
else
//display invalid
cout<<"Invalid!";
cout<<endl;
//return the value
return 0;
}
Learn more grading system
brainly.com/question/24298916
Answer:
Time Complexity of Problem - O(n)
Explanation:
When n= 1024 time taken is t. on a particular computer.
When computer is 8 times faster in same time t , n can be equal to 8192. It means on increasing processing speed input grows linearly.
When computer is 8 times slow then with same time t , n will be 128 which is (1/8)th time 1024.
It means with increase in processing speed by x factor time taken will decrease by (1/x) factor. Or input size can be increased by x times. This signifies that time taken by program grows linearly with input size n. Therefore time complexity of problem will be O(n).
If we double the speed of original machine then we can solve problems of size 2n in time t.
Answer:
Question was incomplete and continued the question
For each of the following scenarios, which of these choices would be best? Explain your answer.
BST
Sorted Array
Un-sorted Array
a) The records are guaranteed to arrive already sorted from lowest to highest (i.e., whenever a record is inserted, its key value will always be greater than that of the last record inserted). A total of 1000 inserts will be interspersed with 1000 searches.
b) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1,000,000 insertions are performed, followed by 10 searches.
Explanation:
Answer for a: Un-sorted array or Un-sorted linked list : as mentioned in the question itself that the records are arriving in the sorted order and search will not be O(log n) and insert will be not be O(n).
Answer for b : Un-sorted array or Un-sorted linkedlist : Number of the items to be inserted is already known which is 1,000,000 but it is very high and at the same time search is low. Unsorted array or Unsorted linked list will be best option here.
Answer:
I hope this answer is correct
Explanation:
Internal registers include the instruction register (IR), memory buffer register (MBR), memory data register (MDR), and memory address register (MAR). The instruction register fetches instructions from the program counter (PC) and holds each instruction as it is executed by the processor.
