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svlad2 [7]
4 years ago
15

You have an interface on a router with the IP address of 192.168.192.10 /29. What is the broadcast address the hosts will use on

this LAN?
Computers and Technology
1 answer:
MariettaO [177]4 years ago
5 0

Answer: it is 192.168.192.15

Explanation:

A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.

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The 7-bit ASCII code for the character ‘&’ is: 0100110 An odd parity check bit is now added to this code so 8 bits are trans
Lorico [155]

Answer

First part:

The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.

Second part:

The invalid bit sequence are option a. 01001000 and d. 11100111

Explanation:

Explanation for first part:

In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.

If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.

If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.  

Explanation for second part:

A valid odd parity bit sequence will always have odd number of 1s.

Since in option a and d,  total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.

And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.

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3 years ago
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alexgriva [62]
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What is a project?
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Explanation:

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Explanation:

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