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3 years ago

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Consider the following code snippet: int count = 0; int[][] numarray = new int[2][3]; for (int i = 0; i < 3; i++) { for (int

j = 0; j < 2; j++) { numarray[j][i] = count; count++; } } What is the value of numarray[1][2] after the code snippet is executed?

Computers and Technology

1 answer:

inna [77]3 years ago

7 0

I have never seen this in my life sorry

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4 years ago

Read 2 more answers

How long does it take to send a 8 MiB file from Host A to Host B over a circuit-switched network, assuming: Total link transmiss

levacccp [35]

Answer:

Given values are:

$File size= 8 MiB$

$Total Link Transmission Rate = 44.2 Gbps$

$Users = 10$

$Setup Time=85.1 ms$

We know that

1 MiB = $2^{20}$ = 1048576 bytes

1 MiB = $2^{20}$ x8= 8388608 bits

8MiB= 8388608x8 = 67108864 bits

So,

Total Number of bits = 67108864 bits

Now

Total link transmission rate = 44.2 Gbps

1 Gbps = 1000000000 bps

44.2 Gbps = 44.2 x 1000000000 =44200000000 bps

For FDM Network

Transmission rate for one time slot = $\frac{44200,000,000}{10}$ bits per second

Transmission rate for one time slot = 4420000000 bit per second

Transmission rate for one time slot = $\frac{44200,000,000}{10^{3} .10 }$

in milliseconds

Transmission rate for one time slot in =4420000 bits per millisecond

Now,

Total time taken to transmit 8 MiB of file = $\frac{Total Number of Bits}{Transmission Rate}$

Total time taken to transmit 8 MiB of file = $\frac{67108864}{4420000}$

As setup time is also given So,

Total time = Setup time + Transmission Time

Total time = 85.1 + 15.18

Total time = 100.2 milliseconds

Hence it is the time required to send a 8MiB of file from Host A to Host B.

I hope it will hep you!

6 0

3 years ago