A __________ is directed at the user file at the host where passwords, token passcodes, or biometric templates are stored.

Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac

Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

= 15 bits/sec

10 independent connections are established, during parallel downloads, and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

= 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

4 0

2 years ago

3. Personal Trainer is very concerned about the security and protection of the information they collect in the new information s

Sever21 [200]

Answer:

Check the explanation

Explanation:

in order to make sure that the built into the new systems, the below input and output levels of security control have to be considered for personal trainers staff.

There are:

Network Security

Application security

Physical security

User security

File security

Procedural security

These securities levels are consistent and advantage in building the decisions regarding system security.

There are the following policies applied to each control:

Network Security: Network interface and traffic controlling, encryption of data.

Application Security: Verification of data input/output and update of software proceedings.

Physical Security: Security to servers, computers and physical environment.

User Security: Protection passwords and identify management skills.

File Security: Access Permission allocated to users and editing authorizations given to users.

Procedural Security: Certify safety protection to perform decision-making activities and document shredders.

Finally, consider another main point is depends the new system performance are system response time, throughput, bandwidth time and turnaround time.

7 0

2 years ago

Describe the output when the following code executes in 64-bit mode: .data dividend_hi QWORD 00000108h dividend_lo QWORD 3330002

iragen [17]

Answer:

RAX = 333000h (16 bits with preceding zeros removed)

RDX = 20h (also 16 bits with preceding zeros removed)

Explanation:

The "div" opcode in the assembly language source code is used to divide operands. It accepts a divisor ( the denominator) and divides the content of the AX register. The result is saved in the AX register while the remainder (if any) is saved in the DX register. If the DX register holds any data, the data is replaced with the divisor remnant.

The code above divides the content of the RAX register with the divisor variable and saves the result and remainder in the RAX and RDX respectively.

7 0

2 years ago

Tyler wants to upgrade his very old operating system to the latest mac or microsoft operating system, but he cannot seem to inst

nordsb [41]

Every program has minimum system requirement, so at least His CPU is not good, or the amount of his RAM is too short to run the new version of OS.

6 0

2 years ago

A(n) ______of the audio or video clip makes the content easily accessible to people with hearing disabilities.

defon

subtitles

hope that helped you