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Dmitry [639]
2 years ago
14

Consider the following network: proxy-cache On average there are 35 objects downloaded per second, and the average object size i

s 50,000 bits. What is the traffic intensity on the 2 Mbps access link?
a. 0.035
b. 0.007
c. 0.250
d. 0.875
Computers and Technology
1 answer:
Andrews [41]2 years ago
4 0

Answer:

(d) 0.875 erlangs

Explanation:

In a digital network, the traffic intensity = \dfrac {aL}{R}

where

a = average arrival rate of packets = 35 objects downloaded per second

L is the average packet length = 50,000 bits (e.g. in bits), and

R is the transmission rate = 2 Mbps = 2,000,000 bits per second

Traffic Intensity = \dfrac {aL}{R}=\dfrac {50000 X 35}{2000000}=\dfrac {1700000}{2000000}=0.875

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Write a program that plays a reverse guessing game with the user. The user thinks of a number between 1 and 10, and the computer
nika2105 [10]

Answer:

  1. import random  
  2. target = 7  
  3. count = 0
  4. for i in range(100):
  5.    guess = random.randint(1,10)  
  6.    
  7.    if(guess == target):
  8.        count += 1
  9. print("Total of correct guess: " + str(count))

Explanation:

The solution is written in Python 3.

Firstly, import the random module since we are going to simulate the random guess by computer. Next, we presume user set a target number 7 (Line 3). Create a counter variable to track the number of correct guess (Line 4).

Presume the computer will attempt one hundred times of guessing and we use randint to repeatedly generate a random integer between 1 - 10 as guess number (Line 6). If the guess number is equal to the target, increment count by one (Line 8-9).

Display the total number of right guess to terminal (Line 11).  

4 0
2 years ago
George and Miguel are considering opening up a shoe store but first need to do market research. Which one of these is NOT part o
Morgarella [4.7K]
The correct answer is D, Draw conclusions and make decisions for their business based on the research results.
7 0
2 years ago
Read 2 more answers
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Molodets [167]

Explanation:

FP - 50 \times 10^6

CPI - 1

INT -110 \times 10^6,

CPI - 1

I/S - 80 \times 10^6 ,

CPI - 4

Branch - 16 \times 10^6

CPI - 2

Clock Speed - 2 \times 10^9

Time(old) =\frac{50 x 10^6 + 110 x 10^6 + 4 x ( 80 x 10^6) + 2 x (16 x 10^ 6)}{2 x 10^9}

Time(old) = 256 \times 10^ {-3}

Time(new) =  \frac{256 \times 10^{-3}}{2}

                = 128 \times 10^{-3}

                =\frac{CPI(new) x [50 x 10^6 + 110 x 10^6 + 4 x ( 80 x 10^6) + 2 x (16 x 10^ 6)]}{2 x 10^9}

                =  128 \times 10^{-3}

CPI(new) = \frac{-206}{50}

               = -4.12

3 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
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tigry1 [53]
Email client as port 25 is the default port for an email server.
6 0
2 years ago
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