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3 years ago

5

Gps receivers are commonly used by individuals to determine their geographic location while hiking and to obtain driving directi

ons while traveling.
a. True
b. False

Computers and Technology

1 answer:

soldier1979 [14.2K]3 years ago

7 0

I think it is true also.

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Consider the latency model for static congestion windows. If the server receives an acknowledgement for the first data segment i

Angelina_Jolie [31]

Answer:

Latency of an object O is shown below.

Explanation:

W segment and stalled state transmits nothing and waits for acknowledgement. The latency is 2 R TT + the time required for server that are using to transmit the object + the amount of time when server is in stalled state. Let K be the number of window that is K= O/WS .The serveries stalled state where K-1 is period of ime with period lasting RTT-(W-1)S/R

latency = 2RTT +O/R +(K-1)[S/R +RTT - WS/R]

After combining the two case

latency = 2 RTT + O/R + (K-1)[S/R +RTT - WS/R]

where [x] means maximum of (x.0). This is the complete ananlysis of the static windows.

server time for transmit the object is (K-1)[S/R +RTT - WS/R]

7 0

3 years ago

What is segmentation and paging?

iogann1982 [59]

Over the years, operating systems have sought to be more efficient, which is why it is vital that the use of main memory such as ram be as intelligent as possible, so that operating systems are more efficient.

Segmentation is a process of dividing the program into logical units, such as sub functions, arrays, variables, etc., making it possible to have processes divided into pieces so that it is easy to access each of the processes that this leads to its execution. Segmentation allows the programmer to contemplate the memory as if it had several address spaces or segments. References to memory consist of an address of the form segment number - offset.

Pagination is a technique where memory space is divided into physical sections of equal size, called page frames. The programs are divided into logical units, called pages, that are the same size as the page frames. In this way, an information page can be loaded in any page frame. The pages serve as an information storage unit and transfer between main memory and auxiliary or secondary memory. Each frame is identified by the frame address, which is in the physical position of the first word in the page frame.

8 0

3 years ago

Josie, your best friend, is having a huge Sweet Sixteen party. She has designated you to address all 100 of the envelopes (using

marysya [2.9K]

You add all the addresses into excel and and do mail-merge with the invitation. Adding the 100 addresses is probably a lengthy process but once this is complete you can just add on addresses and once a year or whenever you have to mail something to everyone, it's a only mouse-click.

4 0

3 years ago

Convert the following numbers. (Please show all steps; no marks will be awarded if no steps are shown) [1.5 x 4 = 6 marks]

katen-ka-za [31]

Answer:

i. 10210212

ii. 100 101 001 in BASE 2

iii. 46.4631

iv. 12.453125

Explanation:

i. Converting to base 10 we get

= 10 x 162 + 13 x 161 + 9 x 160 = 2777

Converting to base 10 we get =

2777/3 = 925 with remainder 2

925/3 = 308 with remainder 1

308/3 = 102 with remainder 2

102/3 = 34 with remainder 0

34/3 = 11 with remainder 1

11/3 = 3 with remainder 2

3/3 = 1 with remainder 0

1/3 = 0 with remainder 1

Hence the Answer is 10210212

ii. = Octal is taken in pair of 3 bits , Writing binary of each number in 3 bits we get = 100 101 001 in BASE 2

iii. = 1 x 52 + 2 x 51 + 3 x 50 + 3 x 5-1 = 38.6 in decimal

Converting to Octal we get

38 /8 = 4 with remainder 6

4 /8 = 0 with remainder 4

.6 x 8 = 4.8

.8 x 8 = 6.4

46.4631

iv. = 1x 81 + 4 x 80 + 3 x 8-1 + 5 x 8-2 = 12.453125

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4 0

3 years ago

Use ordinary pipes to implement an inter-process communication scheme for message passing between processes. Assume that there a

Naya [18.7K]

Answer:

Yup

Explanation:

6 0

3 years ago