Answer:
10 20
30 40
50 60
Explanation:
Given
The above code segment
Required
What is printed, if anything
To do this, we analyze the code line by line.
Line 1: The first line creates a 4 by 3 array named mat
Line 2: for (int[] row : mat)
{ -> This creates row[] array which represents each row of array mat
Line 3: for (int j = 0; j < row.length; j += 2){ -> This iterates through the even indexed elements of the row array i.e. 0 and 2
Line 4: System.out.print(row[i] + " "); -> This prints the even indexed elements of the row array.
The even indexed elements are: 10, 20, 30, 40, 50 and 60
Line 5: System.out.println();
--> This prints a new line
A band from the 1970s has reunited. They want to revive their careers, but they do not have the funding they once did. An international tour is not financially possible. The band needs to build renewed interest in their music and make a profit.
Which of the following would be the best decision for this band?
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Correct Answer:
<u><em>B) Plan a one-night reunion concert that will be broadcast live, online, and on cable, then sell recordings of the event.</em></u>
<u><em></em></u>
<em>Explanation:</em>
I got the answer correct on EDG2020.
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Can you post the work assigned so I can better understand?
Answer:
131 packets
Explanation:
Given
W (link ) =0.5 M bits, and Ping packets ( P ) = 500bytes
To flood a target organization that has a bandwidth of W bits using a packet Mbits, you will require a W / P packet
P ( required ) = 500 * 8 = 4000 bits
therefore the W /P packet is calculated as
= 0.5 M bits / 4000 bits
given that : 1 M bits = 1,048,576 bits
0.5 M bits / 4000 bits= 131.07
hence approximately 131 packets/ sec is required to flood the target organization
Answer:
Under the armpit, which is called the axillary method, and also through mouth, which is called the oral method.
Explanation:
