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3 years ago

14

In a penetration test, a ________ team consists of IT staff who defend against the penetration testers. They are generally aware

that a penetration test is happening, but do not know what methods the testers will use.

Computers and Technology

1 answer:

Crazy boy [7]3 years ago

7 0

Answer: BLUE team

Explanation:

A penetration test is usually performed because it is used to know the defense mechanism of the system against hackers and other unethical access to the system.

In order to defend the system from the penetration testers we have the blue team whose function is to develop algorithms and identify all the security flaws in the system so that it is not under attack from hackers.

The Blue team are generally aware o the penetration test however they do not know what mechanisms or methods the penetration team will be using.

So system upon passing the penetration test can be considered safe from outside attacks.

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g Create a program that prompts a user to enter a login name and password. The correct login name is Admin, and the correct pass

frutty [35]

Answer:

Answered below.

Explanation:

//Program in Python

login_name = "Admin"

password = "PASS"

display = " "

user_login_name = input ("Enter username: ")

user_password = input("Enter your Password: ")

if user_login_name == login_name and user_password == password:

display = "Correct Login"

elif user_login_name != login_name:

display = "wrong username"

elif user_password != password:

display = "wrong password"

print(display)

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3 years ago

Joe is examining the logs for his web server and discovers that a user sent input to a web application that contained the string

alexira [117]

SQL Injection Attack

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2 years ago

Note: You can use a word document to write your answers and copy-paste your answer to the area specified. a. (5 points) Convert

MrMuchimi

Answer:

$EA9_{16} = 3753$

$CB2_{16} = 3250$

$(1011 1110 1101 1011 1010)_2 = 781754$

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

$(1011 1110 1101 1011 1010)_2 = BEDBA$

$(1010 1000 1011 1000 1110 1101)_2 = A8B8ED$

$74510_8= 221416$

$67210_8 = 203212$

Explanation:

Solving (a): To base 10

$(i)\ EA9_{16$

We simply multiply each digit by a base of 16 to the power of their position.

i.e.

$EA9_{16} = E * 16^2 + A * 16^1 + 9 * 16^0$

$EA9_{16} = E * 256 + A * 16 + 9 * 1$

In hexadecimal

$A = 10; E = 14$

So:

$EA9_{16} = 14 * 256 + 10 * 16 + 9 * 1$

$EA9_{16} = 3753$

$(ii)\ CB2_{16}$

This gives:

$CB2_{16} = C * 16^2 + B * 16^1 + 2 * 16^0$

$CB2_{16} = C * 256 + B * 16 + 2 * 1$

In hexadecimal

$C = 12; B =11$

So:

$CB2_{16} = 12 * 256 + 11 * 16 + 2 * 1$

$CB2_{16} = 3250$

Solving (b): To base 10

$(i)\ (1011 1110 1101 1011 1010)_2$

We simply multiply each digit by a base of 2 to the power of their position.

i.e.

$(1011 1110 1101 1011 1010)_2 = 1 * 2^{19} + 0 * 2^{18} + 1 * 2^{17} + 1 * 2^{16} +1 * 2^{15} + 1 * 2^{14} + 1 * 2^{13} + 0 * 2^{12} + 1 * 2^{11} + 1 * 2^{10} + 0 * 2^9 + 1 * 2^8 +1 * 2^7 + 0 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 0 * 2^0$

$(1011 1110 1101 1011 1010)_2 = 781754$

$(ii)\ (1010 1000 1011 1000 1110 1101)_2$

$(1010 1000 1011 1000 1110 1101)_2 = 1 * 2^{23} + 0 * 2^{22} + 1 * 2^{21} + 0 * 2^{20} +1 * 2^{19} + 0 * 2^{18} + 0 * 2^{17} + 0 * 2^{16} + 1 * 2^{15} + 0 * 2^{14} + 1 * 2^{13} + 1 * 2^{12} +1 * 2^{11} + 0 * 2^{10} + 0 * 2^9 + 0 * 2^8 + 1 * 2^7 + 1 * 2^6 + 1 * 2^5 + 0 * 2^4 + 1*2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0$

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

Solving (c): To base 16

$i.\ (1011 1110 1101 1011 1010)_2$

First, convert to base 10

In (b)

$(1011 1110 1101 1011 1010)_2 = 781754$

Next, is to divide 781754 by 16 and keep track of the remainder

$781754/16\ |\ 48859\ R\ 10$

$48859/16\ |\ 3053\ R\ 11$

$3053/16\ |\ 190\ R\ 13$

$190/16\ |\ 11\ R\ 14$

$11/16\ |\ 0\ R\ 11$

Write out the remainder from bottom to top

$(11)(14)(13)(11)(10)$

In hexadecimal

$A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.$

$(11)(14)(13)(11)(10)=BEDBA$

So:

$(1011 1110 1101 1011 1010)_2 = BEDBA$

$ii.\ (1010 1000 1011 1000 1110 1101)_2$

In b

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

Next, is to divide 11057389 by 16 and keep track of the remainder

$11057389/16\ |\ 691086\ R\ 13$

$691086/16\ |\ 43192\ R\ 14$

$43192/16\ |\ 2699\ R\ 8$

$2699/16\ |\ 168\ R\ 11$

$168/16\ |\ 10\ R\ 8$

$10/16\ |\ 0\ R\ 10$

Write out the remainder from bottom to top

$(10)8(11)8(14)(13)$

In hexadecimal

$A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.$

$(10)8(11)8(14)(13) = A8B8ED$

So:

$(1010 1000 1011 1000 1110 1101)_2 = A8B8ED$

Solving (d): To octal

$(i.)\ 74510$

Divide 74510 by 8 and keep track of the remainder

$74510/8\ |\ 9313\ R\ 6$

$9313/8\ |\ 1164\ R\ 1$

$1164/8\ |\ 145\ R\ 4$

$145/8\ |\ 18\ R\ 1$

$18/8\ |\ 2\ R\ 2$

$2/8\ |\ 0\ R\ 2$

Write out the remainder from bottom to top

$74510_8= 221416$

$(ii.)\ 67210$

Divide 67210 by 8 and keep track of the remainder

$67210/8\ |\ 8401\ R\ 2$

$8401/8\ |\ 1050\ R\ 1$

$1050/8\ |\ 131\ R\ 2$

$131/8\ |\ 16\ R\ 3$

$16/8\ |\ 2\ R\ 0$

$2/8\ |\ 0\ R\ 2$

Write out the remainder from bottom to top

$67210_8 = 203212$

8 0

3 years ago

Which of the following statements is true?

jeka57 [31]

Answer: Constructors can specify parameters but not return types.

Explanation:

public class Student {

int roll_no;

public Student(int a) {

roll_no = a;

}

public static void main(String[] args) {

Student abs = new Student(10);

System.out.println(abc.roll_no);

}

}

In the above code we have illustrated the working of constructors. We have a class with the name Student. then a constructor is created of the class called as the class constructor. In the main we create an object of the class and with this object we invoke the constructor and also pass a parameter. Here in the code we are passing the roll no of the student.

So we can say that constructor is called during the runtime when the object created invokes the constructor so a constructor can have many arguments but it does not have a return type.

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3 years ago

Which number on the image above correctly indicates the name of a folder in this URL?

Anna35 [415]

The answer is definitely C. 3

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3 years ago