Answer:
The algorithm:
Input days
sum = 0
for i = 1 to 
input text
sum = sum + text
end for
average = sum/days
print average
The program in pascal:
var days, sum, text, i:integer;
var average : real;
Begin
write ('Days: '); readln(days);
sum:=0;
for i := 1 to do 
write ('Text: '); readln(text);
sum:=sum+text;
end;
average := (sum/days);
writeln ('The average text is' , average);
End.
Explanation:
This declares all variables
var days, sum, text, i:integer;
var average : real;
This begins the program
Begin
This gets the number of days from the user
write ('Days: '); readln(days);
Initialize sum to 0
sum:=0;
This iterates through the days
for i := 1 to do begin
This gets the text for each day
write ('Text: '); readln(text);
This sums up the texts
sum:=sum+text;
End loop
end;
Calculate average
average := (sum/days);
Print average
writeln ('The average text is' , average);
End program
End.
Answer:
B. #
Explanation:
Every preprocessing directive must begin with the # character.
For example:
#define : Used to define a macro
#ifndef : Conditional evaluation of macro
#include : Used to include other preprocessor header file as part of the code
As we can see each of the directives begins with the # character. This is a signal to the preprocessor to interpret the subsequent keyword as a directive.
Answer: A. Verify the wireless connection is using the fastest wireless standard the router supports.
D. Enable QoS for the gaming applications on the router and on the son's computer.
Explanation: Verifying the wireless connection is using the fastest wireless standard the router supports will improve line of connectivity and reduce network latency hence gaming consoles communicate faster.
QoS enables the prioritizing of the volume of internet traffic on the console for high quality gaming. QoS will also enable more internet bandwidth to be channeled to the gaming console.
Explanation:
A.)
we have two machines M1 and M2
cpi stands for clocks per instruction.
to get cpi for machine 1:
= we multiply frequencies with their corresponding M1 cycles and add everything up
50/100 x 1 = 0.5
20/100 x 2 = 0.4
30/100 x 3 = 0.9
CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8
We find CPI for machine 2
we use the same formula we used for 1 above
50/100 x 2 = 1
20/100 x 3 = 0.6
30/100 x 4 = 1.2
CPI for m2 = 1 + 0.6 + 1.2 = 2.8
B.)
CPU execution time for m1 and m2
this is calculated by using the formula;
I * CPI/clock cycle time
execution time for A:
= I * 1.8/60X10⁶
= I x 30 nsec
execution time b:
I x 2.8/80x10⁶
= I x 35 nsec
