Answer: There are many different types of user interfaces. To decide on the user interface depends entirely on the requirement of the client.
Explanation:
There are different types of interfaces such as command line user interface, graphical user interface, menu based, form based. Therefore to choose among them it depend on the requirement specified by a client. Mostly nowadays GUI is used. to maintain records form based is preferred. For system software CUI is better due to decrease its pressure on the processor. Networking is also both GUI and CUI. So it depend mainly on the type of application developed , client requirements, power consumption based on its dependence on processor power.
It would be much better if you've provided additional information, as it's not clear what your question is about. Anyway, I've found this question with options.
And the answer is Without exception, you must always stop when <span> A traffic officer instructs you to stop</span>.
Because most people really don't want or need to see what's in there.
If you want to, open the door during the cycle. It won't harm anything.
If you're curious to see a dishwasher in action, some appliance showrooms have a display model, where the spray parts and the racks are in an entirely transparent box. They're to show how powerfully and completely the stuff inside will get cleaned, but you can also see how the dishwasher is intended to work, by filling with just a couple of inches of water, and then recirculating it for the duration of the cycle, to loosen the stuff on the dishes.
Answer:
See the code below and the algorithm explanation on the figure.
Explanation:
The explanation in order to get the answer is given on the figure below.
Solving this problem with C. The program is given below:
#include <stdio.h>
int main(void) {
int n, Even=0, Odd=0, Zeros=0;
for (;;) {
printf("\nEnter the value the value that you want to check(remember just integers): ");
//IF we input a non-numeric character the code end;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
Zeros++;
}
else {
if (n % 2) {
Even++;
}
else {
Odd++;
}
}
}
printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);
return 0;
}
