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Allushta [10]
2 years ago
11

The fact that the speed of a vehicle is lower than the prescribed limits shall relieve the driver from the duty to decrease spee

d when approaching and crossing an intersection. True or false
Computers and Technology
2 answers:
blsea [12.9K]2 years ago
4 0

Explanation:

the answer is false ........

Harrizon [31]2 years ago
4 0

Answer:

Of course. ......false

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Which method allows a computer to react accordingly when it requests data from a server and the server takes too long to respond
Rina8888 [55]

Answer:

A. Request timeout.

Explanation:

The end devices like the computer systems in a network seeks to share resources with one another and/ or request resources from central server.

With this, there are two ways computers in a network can communicate. They are peer to peer network communication and client-server network communication.

The client-server communication requires a dedicated central server where computers in the network require data. Peer to peer describes a network where computers serve as both client and server to each other.

Request timeout is a message sent to a source when the time to live period (TTL) of a packet expires.

6 0
2 years ago
int) You are the head of a division of a big Silicon Valley company and have assigned one of your engineers, Jim, the job of dev
sergeinik [125]

Answer:

Correct option is E

Explanation:

a) 2n^2+2^n operations are required for a text with n words

Thus, number of operations for a text with n=10 words is 2\cdot 10^2+2^{10}=1224 operation

Each operation takes one nanosecond, so we need 1224 nanoseconds for Jim's algorithm

b) If n=50, number of operations required is 2\cdot 50^2+2^{50}\approx 1.12589990681\times 10^{15}

To amount of times required is 1.12589990681\times 10^{15} nanoseconds which is

1125899.90685 seconds (we divided by 10^{9}

As 1$day$=24$hours$=24\times 60$minutes$=24\times 60\times 60$seconds$

The time in seconds, our algortihm runs is \frac{1125899.90685}{24\cdot 60\cdot 60}=13.0312 days

Number of days is {\color{Red} 13.0312}

c) In this case, computing order of number of years is more important than number of years itself

We note that n=100 so that 2(100)^2+2^{100}\approx 1.267650600210\times 10^{30} operation (=time in nanosecond)

Which is 1.267650600210\times 10^{21} seconds

So that the time required is 1.4671881947\times 10^{16} days

Each year comprises of 365 days so the number of years it takes is

\frac{1.4671881947\times 10^{16}}{365}=4.0197\times 10^{13} years

That is, 40.197\times 10^{12}=$Slightly more than $40$ trillion years$

4 0
2 years ago
Determining the Services Running on a Network Alexander Rocco Corporation has multiple OSs running in its many branch offices. B
Sladkaya [172]

Answer: provided in the explanation section

Explanation:

 Network’s vulnerabilities:

           Vulnerability is a weak spot in your network that might be exploited by a security threat. Risks are the potential consequences and impacts of unaddressed vulnerabilities. In other words, failing to do Windows Updates on your Web server is vulnerability.

           Regularly scheduled network vulnerability scanning can help an organization identify weaknesses in their network security before the bad guys can mount an attack. The goal of running a vulnerability scanner or conducting an external vulnerability assessments is to identify devices on your network that are open to known vulnerabilities without actually compromising your systems.

           The overall objective of a Vulnerability Assessment is to scan, investigate, analyze and report on the level of risk associated with any security vulnerabilities discovered on the public, internet-facing devices and to provide your organization with appropriate mitigation strategies to address those discovered vulnerabilities.

         

Network topology issues:

Coverage Topology

           Coverage problem reflects how well an area is monitored or tracked. The coverage and connectivity problems in networks have received considerable attention in the research community in recent years

 

Geographic routing

           Geographic routing uses geographic and topological information of the network to achieve optimal routing schemes with high routing efficiency and low power consumption

Port Numbers

           Port Numbers While IP addresses determine the physical endpoints of a network connection, port numbers determine the logical endpoints of the connection. Port numbers are 16-bit integers with a useful range from 1 to 65535.

Port numbers are assigned by an organization called IANA and ports are allocated to various needs to avoid confusion.

Ports are classified into 3 main categories.

Well Known Ports (Port numbers 0 - 1023)

           In a client-server application, the server usually provides its service on a well-known port number. Well-known port numbers are a subset of the numbers which are assigned to applications. According to RFC1700 [5], well-known port numbers are managed by the Internet Assigned Numbers Authority (IANA). They used to be in the range from 1 to 255, but in 1992 the range was increased up to 1023.

Registered Ports (Port numbers1024 - 49151)

Such ports are used by programs run by users in the system.

           In addition to the well-known ports below 1024 there are more port numbers assigned to applications but are located anywhere from 1024 to 65535.

Private or Dynamic Ports (Port numbers 49152 - 65535)

           Private ports are not assigned for any specific purpose.

         

Discuss the concepts of well-known ports and give a brief description of the most commonly used ports: 20, 21, 23, 25, 53, and 110.

Commonly Used Port Numbers

The following port numbers are unofficial list of commonly used for linux/unix based servers.

         

20&21               TCP                FTP (File server protocol)

           FTP is one of the most commonly used file transfer protocols on the Internet and within private networks. An FTP server can easily be set up with little networking knowledge and provides the ability to easily relocate files from one system to another. FTP control is handled on TCP port 21 and its data transfer can use TCP port 20 as well as dynamic ports depending on the specific configuration.

23                   TCP/UDP          Telnet

           Telnet is the primary method used to manage network devices at the command level. Unlike SSH which provides a secure connection, Telnet does not, it simply provides a basic unsecured connection. Many lower level network devices support Telnet and not SSH as it required some additional processing. Caution should be used when connecting to a device using Telnet over a public network as the login credentials will be transmitted in the clear.

25                   TCP/UDP         SMTP   (for sending outgoing emails)

           SMTP is used for two primary functions, it is used to transfer mail (email) from source to destination between mail servers and it is used by end users to send email to a mail system.

53         TCP/UDP        DNS Server (Domain name service for DNS request)

           The DNS is used widely on the public internet and on private networks to translate domain names into IP addresses, typically for network routing. DNS is hieratical with main root servers that contain databases that list the managers of high level Top Level Domains (TLD) (such as .com). T

110             TCP                 POP3 (For receiving emails)

           POP version 3 is one of the two main protocols used to retrieve mail from a server. POP was designed to be very simple by allowing a client to retrieve the complete contents of a server mailbox and then deleting the contents from the server.

4 0
2 years ago
Consider a disk that rotates at 3600 rpm. The seek time to move the head between adjacent tracks is 2 ms. There are 32 sectors p
jasenka [17]

Answer:

19.71 ms

Explanation:

The disk rotates at 3600 rpm, hence the time for one revolution = 60 / 3600 rpm = 16.67 ms.

Hence time to read or write on a sector = time for one revolution / number of sectors per track = 16.67 ms / 32 sectors = 0.52 ms

Head movement time from track 8 to track  9 = seek time = 2 ms

rotation time to head up sector 1 on track 8 to sector 1 on track 9 = 16.67 * 31/32 = 16.15 ms

The total time = sector read time +head movement time + rotational delay + sector write time = 0.52 ms + 2 ms + 16.15 ms + 0.52 ms = 19.19 ms

3 0
2 years ago
Write a static method named countLastDigits that accepts an array of integers as a parameter and examines its elements to determ
Crazy boy [7]

Answer:

See explaination for the program code

Explanation:

code:

public static int[] countLastDigits(int[] list) {

int[] count = new int[10];

for (int i = 0; i < list.length; i++) {

int digit = list[i] % 10;

count[digit]++;

}

return count;

}

8 0
2 years ago
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