Wheels should be pointing towards the curb. this is in case your vehicle rolls, if it does, the wheel will hit the curb and stop the car, it will also prevent the car from going into the road and incoming traffic. vice versa when parking uphill, point wheels away from curb, that is also to prevent the car from rolling to incoming traffic.
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
All of the above are Facts.
preventive maintenance does indeed help with the lifespan etc.
If you are feeling tired or fatigued when operating machinery, stop and rest for fear of mistake. So , two is also correct
4if you dress inappropriately it may cause danger from getting caught in machinery or catching a flame depending on the worksite. so four is correct. Newer models are most likely more updated and have different operations then the last machine. so yes read the manual before using it. so yet again 5 is also correct.
hopefully this helped
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
