The output of the line System.out.println(grid[3][2]) will be: 11.
In the given code snippet, a two-dimensional array named grid is declared with dimensions 7 rows and 4 columns. The code then initializes the elements of the array using a nested loop.
The loop iterates over each row (i) and then each column (j) within that row.
The variable c is used to assign a value to each element of the array in a sequential manner, starting from 0 and incrementing by 1 with each iteration of the inner loop.
When the line of code System.out.println(grid[3][2]); is executed, it accesses the element at row 3 and column 2 of the grid array.
Since arrays in Java are zero-indexed, the row and column indices start from 0. In this case, grid[3][2] refers to the element in the fourth row (index 3) and the third column (index 2) of the array.
Considering the given loop logic, the elements of the grid array are assigned values in a sequential manner.
Therefore, when System.out.println(grid[3][2]) is executed, it will display the value that corresponds to the element at the intersection of the fourth row and the third column of the grid array.
Given that the elements are assigned values starting from 0 and incrementing sequentially, the value displayed will be the third value in the fourth row of the array, which is 11.
Therefore, the output of the line System.out.println(grid[3][2]) will be: 11.
This assumes that no modifications are made to the array between its initialization and the execution of the System.out.println statement.
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assuming stp and a stoichiometric amount of nh3 and no in an expandable container originally at 15 L, what is the final volume if the reaction goes to completion? 4NH3 (g) + 6NO (g) —> 5N2 (g) + 6H2O (g)
Answer:
16.5 L
Explanation:
We are given the following factors (they are all essential):
4 moles NH3 gas (15 L initially)6 moles NO gasContainer volume = 15 L initially at STPGiven reaction:
4NH3 (g) + 6NO (g) -> 5N2 (g) + 6H2O (g)
Using the mole ratios from the balanced chemical equation:
For every 4 moles NH3, there will be 6 moles NO and they will react to produce:
5 moles N26 moles H2OInitial moles of gas:
4 moles NH36 moles NOTotal = 10 molesFinal moles of gas:
5 moles N26 moles H2OTotal = 11 molesBy Avogadro's law, volume is directly proportional to moles of gas at constant temperature and pressure.
Since we have 11 moles of gas after the reaction compared to 10 moles initially, the final volume will be:
(11 moles gas)(15 L initial)/(10 moles initial)
= 16.5 L
So in summary, the final volume of the gas mixture after the reaction goes to completion will be 16.5 L, assuming a constant temperature and pressure within the expandable container.
The general approach is to calculate the mole ratios and number of moles of gases before and after the reaction, then use proportionality to determine how the volume will change accordingly.
On January 1, 2021, Frontier World issues $40.0 million of 8% bonds, due in 15 years, with interest payable semiannually on June 30 and December 31 each year. The proceeds will be used to build a new ride that combines a roller coaster, a water ride, a dark tunnel, and the great smell of outdoor barbeque, all in one ride. 3-a. If the market rate is 9%, calculate the issue price. (FV of $1, PV of $1, FVA of $1, and PVA of $1) (Use appropriate factor(s) from the tables provided. Do not round interest rate factors. Enter your answers in dollars not in millions. Round "Market interest rate" to 1 decimal place. Round your final answers to the nearest whole dollar.)
Choose the answer that best completes the visual analogy.
Answer: "Look at attachment"
Explanation:
If we look at the second representation of the first visual analogy, we see that the star is in the middle of the white circle and is black. Therefore, for the second representation of the second analogy, the circle must be black and in the middle like the black star.
The answer that best completes the visual analogy is option C.
What is the best visual analogy?The answer that ebst completes the visual analogy must be one that follows the same principles as the ones in the first instance. In the first instance, we can see that the star is now in the middle of the circle and is also a dark color.
The second circle is white and the outermost one is dark. Only the third option matches this depiction.
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