You actually have the correct answer. An excel formula starts with = so your answer would be =4*6
Answer:
answer:
#include <iostream>
#include<list>
using namespace std;
bool Greater(int x) { return x>3; } int main() { list<int>l; /*Declare the list of integers*/ l.push_back(5); l.push_back(6); /*Insert 5 and 6 at the end of list*/ l.push_front(1); l.push_front(2); /*Insert 1 and 2 in front of the list*/ list<int>::iterator it = l.begin(); advance(it, 2); l.insert(it, 4); /*Insert 4 at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl; l.erase(it); /*Delete the element 4 inserted at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl;
l.remove_if(Greater); for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " ";
/*Display the list*/
cout<<endl; return 0;
}
Technician A is correct.
Electrochemical batteries can be classified into two main categories; Primary Cells and Secondary Cells.
A primary battery or cell cannot easily be recharged after use and is usually recommended to be discarded following discharge. Most primary cells are termed dry due to their capability of utilizing electrolytes contained within absorbent material.
A secondary cell, on the other hand, can be recharged electrically to their original pre-discharge condition.
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
