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3 years ago

7

3. A vulnerability is: a. A hacker searching for open ports b. A known attack method c. An incorrectly implemented policy d. All

of the above e. None of the above

Computers and Technology

1 answer:

ELEN [110]3 years ago

7 0

Answer:

A

Explanation:

A hacker searching for open ports denotes vulnerability of computer (to hacking).

Cheers

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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)

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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

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

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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

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∞ n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

n-1

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Answer:

class Student(object):

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for count in range(number):

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return max(self.scores)

def __eq__(self,student):

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def __ge__(self,student):

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def __lt__(self,student):

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def __str__(self):

return "Name: " + self.name + "\nScores: " + \

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def main():

student = Student("Ken", 5)

print(student)

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print(student)

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student2 = Student("Ken", 5)

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print(student.__eq__(student2))

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print(student.__eq__(student3))

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3 years ago