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TEA [102]
3 years ago
10

PLEASE SOMEONE PLEASE HELP ME OUT!!!!

Computers and Technology
2 answers:
Arlecino [84]3 years ago
7 0

Answer:

the 1st one

Explanation:

alina1380 [7]3 years ago
6 0

Answer:   Ideate

Explanation:

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The ____ project was developed to make information widely available in an attempt to thwart internet and network hackers.
stira [4]

The Honeynet project existed and developed to create information widely available in an attempt to thwart internet and network hackers.

<h3>What is Honeynet?</h3>

The Honeynet Project exists as an international security research organization, "dedicated to exploring the latest attacks, developing open source security tools to enhance Internet security and learning how hackers behave.

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The Honeynet Project concentrates on three primary goals: Raise attention to the existing threats on the Internet. Conduct research covering data research approaches, unique security tool development, and gathering data about attackers and malicious software they utilize.

Hence, The Honeynet project existed and developed to create information widely available in an attempt to thwart internet and network hackers.

To learn more about Honeynet refer to/;

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2 years ago
The presentation ____ determines the formatting characteristics of fonts and colors.
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The presentation theme determines the formatting characteristics of fonts and colors.
4 0
4 years ago
in cell h5, enter a formula that will calculate the percentage of attendees that went to the Altamonte springs job fair in 2018.
xz_007 [3.2K]
Formula is 

 = (a/b)(100)
8 0
3 years ago
Read 2 more answers
Write a C++ program that reads students' names followed by their test scores. The program should output each students' name foll
Mashutka [201]

Answer:

#include<iostream>

#include<conio.h>

using namespace std;

struct studentdata{

char Fname[50];

char Lname[50];

int marks;

char grade;

};

main()

{

studentdata s[20];

for (int i=0; i<20;i++)

   {

cout<<"\nenter the First name of student";

cin>>s[i].Fname;

cout<<"\nenter the Last name of student";

cin>>s[i].Lname;

cout<<"\nenter the marks of student";

cin>>s[i].marks;

}  

 

for (int j=0;j<20;j++)

{

if (s[j].marks>90)

{

 s[j].grade ='A';

}

else if (s[j].marks>75 && s[j].marks<90)

{

   s[j].grade ='B';

}

else if (s[j].marks>60 && s[j].marks<75)

{

 s[j].grade ='C';

}

else

{

 s[j].grade ='F';

}

}

int highest=0;

int z=0;

for (int k=0;k<20; k++)  

{

if (highest<s[k].marks)

{

 highest = s[k].marks;

 z=k;

}

 

}

cout<<"\nStudents having highest marks"<<endl;

 

cout<<"Student Name"<< s[z].Fname<<s[z].Lname<<endl;

cout<<"Marks"<<s[z].marks<<endl;

cout<<"Grade"<<s[z].grade;

getch();  

}

Explanation:

This program is used to enter the information of 20 students that includes, their first name, last name and marks obtained out of 100.

The program will compute the grades of the students that may be A,B, C, and F. If marks are greater than 90, grade is A, If marks are greater than 75 and less than 90, grade is B. For Mark from 60 to 75 the grade is C and below 60 grade is F.

The program will further found the highest marks and than display the First name, last name, marks and grade of student who have highest marks.

6 0
3 years ago
NAND is logically complete. Use only NAND gates to constructgate-level circuits that compute the
Drupady [299]

Answer:

Hi, for this exercise we have two laws to bear in mind:

Morgan's laws

NOT(А).NOT(В) = NOT(A) + NOT (B)

NOT(A) + NOT (B) = NOT(А).NOT(В)

And the table of the Nand

INPUT OUTPUT

A B A NAND B

0 0         1

0 1         1

1 0         1

1 1         0

Let's start!

a.

Input            OUTPUT

A       A     A NAND A

1         1             0

0        0            1

b.

Input            OUTPUT

A       B     (A NAND B ) NAND (A NAND B )

0         0            0

0         1            0

1          0             0

1          1             1

C.

Input            OUTPUT

A       B     (A NAND A ) NAND (B NAND B )

0        0           0

0        1             1

1         0            1

1          1            1

Explanation:

In the first one, we only need one input in this case A and comparing with the truth table we have the not gate

In the second case, we have to negate the AND an as we know how to build a not, we only have to make a nand in the two inputs (A, B) and the make another nand with that output.

In the third case we have that the OR is A + B and we know in base of the morgan's law that:

A + B = NOT(NOT(А).NOT(В))

So, we have to negate the two inputs and after make nand with the two inputs negated.

I hope it's help you.

3 0
3 years ago
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