In entry section, the process requests permission to access and modify variables shared with others.
<h3>What is Entry section?</h3>
Entry Section is known to be an aspect of any given process that tells the entry of a specific process.
Note that it often allows one process to enter and alter the shared variable and as such, In entry section, the process requests permission to access and modify variables shared with others.
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An area of improvement that can be achieved using AI and Machine in the Future of Transport are given below:
The technical issues of artificial intelligence are
- Artificial Neural Networks.
- Deep Learning.
- Natural language processing.
The potential issues that can arise from the adoption of AI technology
is hacking of transport system and system fail and how it can be overcome is by aiding system with enough security as well as regular maintenance.
<h3>What is the Future of Transport about?</h3>
AI is known to be one that has the power as well as the potential to be able to make traffic in regards to movement of vehicles to be very more efficient.
Note that it is one that can ease traffic congestion and also gives driver's a lot of time, also make parking easier as well as encourage car or ridesharing.
Note that as AI aids or function to keep road traffic to be flowing, it can also lower the fuel consumption rate that is as a result of the use of vehicles idling if stationary,
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I'm pretty sure you would use A.) Quotation Marks.
Hope I helped!
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 − )
Answer:
#include <iostream>
#include <string>
#include <stack>
#include <math.h>
using namespace std;
int main() {
string s;
double n=0;
int position=0;
stack<int> wholeNumbers;
cout<<"Enter a decimal number:";
cin>>s;
string::iterator counter = s.begin();
while(*counter!='.' && counter!=s.end()){
wholeNumbers.push(*counter-48);
counter++;
position=position+1;
}
for(int i=0;i<position;i++){
n=n+(wholeNumbers.top()*pow(10,i));
wholeNumbers.pop();
}
position=-1;
if(counter!=s.end()){
counter++;
}
while(counter!=s.end()){
n=n+((*counter-48)*pow(10,position));
position=position-1;
counter++;
}
cout<<n;
}
Explanation:
- Inside the while loop, push the push a number to the wholeNumbers stack by subtracting it with 48.
- Increment the counter and position variable by 1 inside the while loop.
- Count the number of digit, push each digit to top of stack and find the end of the number,
- Run a for loop up to the value of position variable and pop a value from the wholeNumbers stack.
