All drivers
All drivers who share the road with an impaired driver are at risk.
Normal or random variations that are considered part of operating the system at its current capability are <u> c. common cause variations.</u>
Explanation:
Common cause variation is fluctuation caused by unknown factors resulting in a steady but random distribution of output around the average of the data.
Common-cause variation is the natural or expected variation in a process.
Common-cause variation is characterised by:
- Phenomena constantly active within the system
- Variation predictable probabilistically
- Irregular variation within a historical experience base
It is a measure of the process potential, or how well the process can perform when special cause variation removed.
Common cause variation arises from external sources that are not inherent in the process and is where statistical quality control methods are most useful.
Statistical process control charts are used when trying to monitor and control 5- and 6-sigma quality levels.
Answer:
B. 1 6 3
Explanation:
Given function definition for calc:
void calc (int a, int& b)
{
int c;
c = a + 2;
a = a * 3;
b = c + a;
}
Function invocation:
x = 1;
y = 2;
z = 3;
calc(x, y);
cout << x << " " << y << " " << z << endl;
- Since x is passed by value, its value remains 1.
- y is passed by reference to the function calc(x,y);
Tracing the function execution:
c=3
a=3
b=c+a = 6;
But b actually corresponds to y. So y=6 after function call.
- Since z is not involved in function call, its value remain 3.
So output: 1 6 3
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:
50%
Explanation:
The markup is the difference between the selling price and the cost price. If the mark up is greater than zero, it means there is a profit, if the markup is less than 0, it means there is a loss and if the markup is equal to 0, it means there is breakeven.
Percentage markup = (markup/cost price) * 100%
Selling price - cost price = markup
15 - cost price = 5
cost price = 10
Percentage markup = (markup/cost price) * 100% = (5/10) * 100% = 50%
