Answer:
The total const is 13025 KWh
Explanation:
These are the steps to solve this problem:
- Convert all the powers from W to KW dividing by 1000.
- Convert all the times on minutes to hours dividing by 60.
- Then you can apply energy consumption formula
for any of the appliances. the results will be at KWh. - Sum all the consumtions and you will have the total cost.
Attached you will have a spreadsheet as a guidance. Any questions, just let me know.
With the aid of pointer-based arithmetic operations and the usage of pointers in comparison operations, address arithmetic is a technique for determining the address of an object. Pointer arithmetic is another name for address arithmetic.
The pointers can be used for mathematical operations like addition, subtraction, etc. The outcome of an arithmetic operation on the pointer, however, will likewise be a pointer if the other operand is of type integer because we know that the pointer includes the address. These operations are addition and subtraction. In C++, a pointer's value can be increased or decreased. It signifies that we can change the pointer's value by adding or removing integer values. A pointer arithmetic can be subtracted (or added) from another in a manner similar to this.
Learn more about arithmetic here-
brainly.com/question/11424589
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Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:a) Mainframes, miniframes, PCs, desktop Internet computing, mobile computing
Explanation: The hierarchy of the computers or computing are in the generation is:-
- Mainframe:-designed for the execution of huge data processing , storage and execution.
- Miniframe:-has similar functioning as the mainframe but on smaller scale and version.
- Personal computer(PC):-It is designed for the individual purpose and according the need of the user.
- Mobile computing:-computing that can be done on the mobile phone similar to the computing on the personal computer
- Internet computing:-the computing of the mobile and computer system with the facility of the internet connectivity.
