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3 years ago

ENG103 DISCUSSION BOARD 5

1 answer:

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Plagiarism is a crime, flat out. It is unethical and immoral to copy, verbatim, another persons work and claim it as your own, especially for scholastic uses. In order to avoid plagiarism, it is important that we give accurate references and paraphrase, rather that copy and paste information. It is important to consider the widespread effect of plagiarism. It effects all involved: The student, the owner of the information, the instructor, and even the site from which the information was stolen. Use the proper channels and be honest when completing assignments goes a long way. We must do what we can to prevent plagiarism, starting with ourselves and our own research. <span />

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WHAT THE DEFINITION OD ENGINEER. No CHEAT

Harrizon [31]

A person who designs, maintains or builds machines.

4 0

3 years ago

Read 2 more answers

Define the method object inc_num_kids() for PersonInfo. inc_num_kids increments the member data num_kids.Sample output for the g

topjm [15]

Answer:

class PersonInfo:

def __init__(self):

self.num_kids = 0

def inc_num_kids(self):

self.num_kids += 1

person1 = PersonInfo()

print('Kids:', person1.num_kids)

person1.inc_num_kids()

print('New baby, kids now:', person1.num_kids)

Explanation:

Line 1 of the code, we define the class PersonInfo. Line 3 of the code is the function that will increment the member data num_kids.

4 0

3 years ago

Which of the following is not one of the steps a company would take in an attempt to prevent a malfunction or failure of their p

bulgar [2K]

Answer:

A overdesign

Explanation:

7 0

3 years ago

1. Write a high level algorithm for cooking a cheeseburger.

pishuonlain [190]

Answer:

1. A high level algorithm for cooking a cheeseburger could be:

Heat fry pan
Cook one side of the hamburger
Wait
Turn hamburger upside down
Put cheese over hamburger
Wait
Cut hamburger bread in half
Put cooked hamburger inside bread
End (eat)

2. A detailed algorithm for cooking a cheeseburger could be:

Place fry pan over the stove heater
Turn on heater (max temp)
IF fry pan not hot: wait, else continue
Place raw hamburger on fry pan
IF hamburger not half cooked: Wait X time then go to line 5, else continue
Turn hamburger upside down
Put N slices of cheese over hamburger
IF hamburger not fully cooked: Wait X time then go to line 8, else continue
Turn off heater
Cut hamburger bread in half horizontally
Put cooked hamburger on one of the bread halves.
Put second bread half on top of hamburger
End (eat)

Explanation:

An algorithm is simply a list of steps to perform a defined action.

On 1, we described the most relevant steps to cook a simple cheeseburger.

Then on point 2, the same steps were taken and expanded with more detailed steps and conditions required to continue executing the following steps.

In computational terms, we used pseudo-code for the algorithm, since this is a list of actions not specific to any programming language.

Also we can say this is a structured programming example due to the sequential nature of the cooking process.

7 0

3 years ago

A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co

tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k ...(1)

probability of 2 hops = k(1-k) ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

= k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

∞ n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

n-1

= 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet = $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

6 0

3 years ago