Answer:
The value of 8n would be 16
8n
8(2)= 16
(uhh this is a math question right? Sorry if it has to deal with tech)
Answer:
1/3
Explanation:
3/18 divided by 2 equals 1/3
hope this helps
have a good day
This can be done in a number of steps.
1. Prepare your email request early.
2. Choose an appropriate professor
3. Address the letter properly.
4. Put "Recommendation for [your name]" as the subject line.
5. Start the first paragraph by stating what you want.
6. Outline your relationship with the professor.
7. Use the third paragraph as an opportunity to hint at what you'd like the professor to say about you.
8. Give the details.
9. Close with information about how you will follow-up.
10. Thank the professor, whether or not s/he writes the letter.
11. Follow through as promised by delivering necessary materials and sending a reminder
12. Take responsibility for checking with the scholarship program, graduate school, or prospective employer before the deadline
1. Thank the professor again.
Answer:
(B) A single public IP address that it can use for NAT
Explanation:
Because the IPV4 IP protocol is still used today, the number of available IP addresses is limited (only 4,294,967,296 addresses in the world), for this reason, the most correct practice is the assignment of a single public IP to those companies that acquire services from an ISP, with some few exceptional cases of companies that own several.
So that the company's addressing can be executed successfully, the use of NATs is enabled, this allows the translation of network addresses, allowing the company to have as many private networks as necessary and that these can be communicated Correctly with the global network, the Internet, through the public IP of the company.
Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
