Answer:
Where are the following? You have to post the full question if you want help.
Explanation:
Answer:
Explanation:
#include <iostream>
using namespace std;
int main ()
{
int responses[30],count[6];
int score = 0;
string resp = " ";
for (int i = 0; i < 30; i++)
{
responses[i] = 0;
}
for (int i = 0; i < 6; i++)
{
count[i,1]=0;
count[i,2]=0;
count[i,0]=0;
}
while ((resp != "Y") && (resp != "y"))
{
for (int i = 0; i < 30; i++)
{
while ((score > 5) || (score < 1))
{
cout << "Student " << (i+1)<< " please enter a value (1-5):";
cin >> score;
}
responses[i] = score;
if((score > 5)||(score<1))
{
if(score==1) count[1]++;
if(score==2) count[2]++;
if(score==3) count[3]++;
if(score==4) count[4]++;
if(score==5) count[5]++;
}
score = 0;
}
cout<< "Response Frequency Percentage"<<endl;;
cout<< " 1 "<<count[1]<<" "<<(count[1]/30)<<"%"<<endl;
cout<< " 2 "<<count[2]<<" "<<(count[2]/30)<<"%"<<endl;
cout<< " 3 "<<count[3]<<" "<<(count[3]/30)<<"%"<<endl;
cout<< " 4 "<<count[4]<<" "<<(count[4]/30)<<"%"<<endl;
cout<< " 5 "<<count[5]<<" "<<(count[5]/30)<<"%"<<endl;
cout<< "Do you want to exit? Press Y to exit any other key to continue: ";
cin>> resp;
}
return 0;
}
Answer:Effective but not efficient
Explanation:
Jim is effective because he was able to complete the IT tasks well but he is not efficient because he didn't submit the result on time because being efficient includes management of time.
The answer that fills in the blank is orientation. It is because orientation does not provide all the details of what they are doing but a guideline of what they must do, in order for them to be guided on the things that they need to learn as they start. They require proper guidance which is the orientation before performing before hand. Especially when they are not familiar on what they need to do.
Answer:
a. 7 bits b. 9 bits c. 1 kB d. 2¹⁶ - 1
Explanation:
a. How many bits are needed for the opcode?
Since there are 72 different operations, we require the number of bits that would contain 72 different operations. So, 2ⁿ ≥ 72
72 = 64 + 8 = 2⁶ + 8
Since n must be an integer value, the closest value of n that would contain 72 different operations is n = 7. So, 2⁷ = 128
So, we require 7 bits for the opcode.
b. How many bits are left for the address part of the instruction?
bits left = bits per word - opcode bit = 16 - 7 = 9 bits
c. What is the maximum allowable size for memory?
Since there are going to be 2⁹ bits to addresses each word and 16 bits for each word, the maximum allowable size for memory is thus 2⁹ × 16 = 512 × 16 = 8192 bits.
We convert this to bytes
8192 bits × 1 byte/8 bits = 1024 bytes = 1 kB
d. What is the largest unsigned binary number that can be accommodated in one word of memory?
Since the number go from 0 to 2¹⁶, the largest unsigned binary number that can be accommodated in one word of memory is thus
2¹⁶ - 1
