f the copyright date on the homepage is more than 1 year old, the website should always be considered unmaintained

Your network administrator finds a virus has been successfully inserted into the network software. The virus itself is now consi

Svetlanka [38]

Answer: Vulnerability of

Explanation:

The network administrator was able to identify that virus before it led to denial of service to users. Since it has been discovered, it is no longer a threat as measures will be taken to eliminate it. But a virus that was successful uploaded shows how vulnerable the system is or how bad the security protections put in place are. Steps has to be taken to ensure it does not reoccur.

7 0

2 years ago

Read 2 more answers

________ is free software created and updated by a worldwide community of programmers. office software a web service cloud-based

Yanka [14]

I would thing ios or apple cloud personally.

6 0

2 years ago

Consider the following skeletal C program: void fun1(void); /* prototype */ void fun2(void); /* prototype */ void fun3(void); /*

natita [175]

Answer:

Check the explanation

Explanation:

a) main calls fun1; fun1 calls fun2; fun2 calls fun3

fun3() d, e, f

fun2() c, d, e

fun1() b, c, d

main() a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

From the above call stack diagram, it is very clear that the last function call is made to fun3().

In fun3(), the local variables "d, e, f" of fun3() will be visible

variable "c" of fun2() will be visible

variable "b" of fun1() will be visible

variable "a" of main() will be visible

b) main calls fun1; fun1 calls fun3

fun3() d, e, f

fun1() b, c, d

main() a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

From the above call stack diagram, it is very clear that the last function call is made to fun3().

In fun3(), the local variables "d, e, f" of fun3() will be visible

variable "b, c" of fun1() will be visible

variable "a" of main() will be visible

c) main calls fun2; fun2 calls fun3; fun3 calls fun1

fun1() b, c, d

fun3() d, e, f

fun2() c, d, e

main() a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

From the above call stack diagram, it is very clear that the last function call is made to fun1().

In fun1(), the local variables "b, c, d" of fun1() will be visible

variable "e, f" of fun3() will be visible

variable "a" of main() will be visible

d) main calls fun1; fun1 calls fun3; fun3 calls fun2

fun2() c, d, e

fun3() d, e, f

fun1() b, c, d,

main() a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

From the above call stack diagram, it is very clear that the last function call is made to fun2().

In fun2(), the local variables "c, d, e" of fun2() will be visible

variable "f" of fun3() will be visible

variable "b" of fun1() will be visible

variable "a" of main() will be visible

The last function called will comprise of all its local variables and the variables other than its local variables from all its preceding function calls till the main function.

8 0

2 years ago

Suppose that the data mining task is to cluster points (with (x, y) representing location) into three clusters, where the points

solong [7]

Answer:

Explanation:

K- is the working procedure:

It takes n no. of predefined cluster as input and data points.

It also randomly initiate n centers of the clusters.

In this case the initial centers are given.

Steps you can follow

Step 1. Find distance of each data points from each centers.

Step 2. Assign each data point to the cluster with whose center is nearest to this data point.

Step 3. After assigning all data points calculate center of the cluster by taking mean of data points in cluster.

repeat above steps until the center in previous iteration and next iteration become same.

A1(4,8), A2(2, 4), A3(1, 7), B1(5, 4), B2(5,7), B3(6, 6), C1(3, 7), C2(7,8)

Centers are X1=A1, X2=B1, X3=C1

A1 will be assigned to cluster1, B1 will be assigned to cluster2 ,C1 will be assigned to cluster3.

Go through the attachment for the solution.

5 0

2 years ago

Write the execution steps if the input is a lion.

balandron [24]

Explanation:

Execution step:

1. Starting the program

2. It will get the animal name

3. Check whether the animal eats only plants. Then it prints "Herbivore". If the condition is false, it prints "carnivore".

4. Stop the program

To identify input, output and decision:

Start and Stop (oval): These are neither input nor output

Does it eat only plants?: Decision making statement ("d")

Read the names of the animals(Parallelogram): Input ("i")

Print Herbivore, Print Carnivore: Output "o"

Scenario 1: Input is lion

1. Start

2. Reads the input as "lion"

3. Check if lion eats only plants. So here it is false

4. Print "Carnivore"

5. Stop

Scenario 1: Input is Elephant

1. Start

2. Reads the input as "Elephant"

3. Check if lion eats only plants. So here it is true

4. Print "Herbivore"

5. Stop

7 0

2 years ago