Answer:
The answer is "Option d".
Explanation:
In the given code, two class "TestA and TestB" is defined, that calculates some values which can be described as follows:
- In class "TestA", three integer variable "x, y, and counter" is declared, that initializes with a value, that is "2, 20, and 0", inside the class for loop is declare that uses variable j which starts from and ends when the value of j is less than 100, it will increment the value of counter variable by 1.
- In the class "TestB", an integer "counter" variable is initializes a value with 0, inside the class the for loop is used that uses variable j, which starts from 10, and ends when j is less than 0. in the loop it increments the value of "counter" variable by 1. that's why in this question except "option d" all were wrong.
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The very first step of the lowest cost method is to find the cell with the lowest cost in the entire matrix representing the cost of transportation along with supply and demand.
C. Find the cell with the lowest cost from the remaining (not crossed out) cells.
<u>Explanation:</u>
The second step in the lowest cost method is to allocate as many units as possible to that cell (having the lowest cost) without exceeding the supply or demand.
Then cross out the row or column (or both) that is exhausted by the assignment made. These two steps are further repeated until all the assignments are made and the total cost of transportation is calculated at the end.
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