Answer:
A total of 79.3mb will be needed
A machine on a 10 Mbps network is regulated by a token bucket algorithm with a fill rate of 3 Mbps. The bucket is initially filled to capacity at 3MB. How long can the machine transmit at the full 10 Mbps capacity
What are the constraints for designing small and large files and how these are resolved in different file system
Answer:
space management and buffering speed.
Explanation:
There are different types of file management systems in a computer system, examples of which are NTFS, FAT, WAFL, etc, and are governed by protocols like NFS, TFTP, FTP, etc.
These file systems are used in storages like the hard disk drive, CD and DVD, solid-state drive, etc, to organize or manage the files from boot setup, device drivers to permission-seasoned user files.
Files in storage range from small to large files, for which the schema of the file system must adjust to manage and allocate free space to other files in the future. The file system is also able to index the location of a file for retrieval to a cache memory, making buffering faster.
(Process scores in a text file) Suppose that a text file contains an unspecified number of scores. Write a program that reads the scores from the file and displays their total and average. Scores are separated by blanks. Your program should prompt the user to enter a filename. Here is a sample run:
Answer:
Here is the Python program:
def scores(file): # method scores that takes a file name as parameter and returns the sum and average of scores in a file
with open(file, 'r') as infile: # open the file in read mode
lines = [score.split() for score in infile] # split the scores into a list
print("The scores are:",lines) #print the scores
for line in lines: # loops through each score
total= sum(int(score) for score in line) # adds the scores
average =total/len(line) # computes average by taking sum of scores and dividing by number of scores in file
print("The sum is:", total) #prints the sum of scores
print("The average is:", "{:.2f}".format(average)) #prints the average
filename = input("Enter name of the file: ") #prompts user to enter name of file
scores(filename) #calls scores method by passing the file name to it in order to compute sum and average of file contents i.e. scores
Explanation:
It is assumed that the scores in the file are separated by a blank space.
The scores() method takes a file name as parameter. Then it opens that input file in read mode using object infile.
split() method is used to split the scores in a file into a list. Suppose the scores are 1 2 3 4 5 6 . So after the split, they become ['1', '2', '3', '4', '5', '6']
The loop iterates through each score in the file, splits them into a list and stores this list in lines. The next print statement prints these scores in a list.
The second loop for line in lines iterates through each score of the list and the statements: total= sum(int(score) for score in line) and average =total/len(line) computes the total and average of scores.
total= sum(int(score) for score in line) statement works as follows:
for loop iterates through each element of list i.e. each score
int() converts that string element into integer.
sum() method adds the integers to compute their total.
So if we have ['1', '2', '3', '4', '5', '6'] each element i.e. 1,2,3,4,5,6 is converted to integer by int() and then added together by sum method. So this becomes 1+2+3+4+5+6 = 21. This result is stored in total. Hence
total = 21.
average = total/len(line) works as follows:
The computed sum of scores stored in total is divided by the number of scores. The number of scores is computed by using len() method which returns the length of the line list. So len() returns 6. Hence
average = total/len(line)
= 21 / 6
average = 3.5
The next two print statement prints the value of sum and average and "{:.2f}".format(average)) prints the value of average up to 2 decimal places.
The screenshot of the program along with its output is attached.
Consider the following calling sequences and assuming that dynamic scoping is used, what variables are visible during execution of the last function called? Include with each visible variable the name of the function in which it was defined.a. Main calls fun1; fun1 calls fun2; fun2 calls fun3b. Main calls fun1; fun1 calls fun3c. Main calls fun2; fun2 calls fun3; fun3 calls fun1d. Main calls fun3; fun3 calls fun1e. Main calls fun1; fun1 calls fun3; fun3 calls fun2f. Main calls fun3; fun3 calls fun2; fun2 calls fun1void fun1(void);void fun2(void);void fun3(void);void main() {Int a,b,c;…}void fun1(void){Int b,c,d;…}void fun2(void){Int c,d,e;…}void fun3(void){Int d,e,f;…}
Answer:
In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.
Explanation:
In main() function three integer type variables are declared: a,b,c
In fun1() three int type variables are declared/defined: b,c,d
In fun2() three int type variables are declared/defined: c,d,e
In fun3() three int type variables are declared/defined: d,e,f
a. Main calls fun1; fun1 calls fun2; fun2 calls fun3
Here the main() calls fun1() which calls fun2() and fun2() calls func3() . This means first the func3() executes, then fun2(), then fun1() and last main()
Visible Variable: d, e, f Defined in: fun3
Visible Variable: c Defined in: fun2 (the variables d and e of fun2
are not visible)
Visible Variable: b Defined in: fun1 ( c and d of func1 are hidden)
Visible Variable: a Defined in: main (b,c are hidden)
b. Main calls fun1; fun1 calls fun3
Here the main() calls fun1, fun1 calls fun3. This means the body of fun3 executes first, then of fun1 and then in last, of main()
Visible Variable: d, e, f Defined in: fun3
Visible Variable: b, c Defined in: fun1 (d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
c. Main calls fun2; fun2 calls fun3; fun3 calls fun1
Here the main() calls fun2, fun2 calls fun3 and fun3 calls fun1. This means the body of fun1 executes first, then of fun3, then fun2 and in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e, f Defined in: fun3 ( d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
Here variables c, d and e of fun2 are not visible
d. Main calls fun3; fun3 calls fun1
Here the main() calls fun3, fun3 calls fun1. This means the body of fun1 executes first, then of fun3 and then in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e, f Defined in: fun3 ( d not visible )
Visible Variable: a Defined in: main (b and c not visible)
e. Main calls fun1; fun1 calls fun3; fun3 calls fun2
Here the main() calls fun1, fun1 calls fun3 and fun3 calls fun2. This means the body of fun2 executes first, then of fun3, then of fun1 and then in last, of main()
Visible Variable: c, d, e Defined in: fun2
Visible Variable: f Defined in: fun3 ( d and e not visible)
Visible Variable: b Defined in: fun1 ( c and d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
f. Main calls fun3; fun3 calls fun2; fun2 calls fun1
Here the main() calls fun3, fun3 calls fun2 and fun2 calls fun1. This means the body of fun1 executes first, then of fun2, then of fun3 and then in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e Defined in: fun2
Visible Variable: f Defined in: fun3
Visible Variable: a Defined in: main
what are the morals and ethics of computer
Answer:
Computer ethics is a part of practical philosophy concerned with how computing professionals should make decisions regarding professional and social conduct. Margaret Anne Pierce, a professor in the Department of Mathematics and Computers at Georgia Southern University has categorized the ethical decisions related to computer technology and usage into three primary influences:
The individual's own personal code.
Any informal code of ethical conduct that exists in the work place.
Exposure to formal codes of ethics.
Explanation:
Cloud computing gives you the ability to expand and reduce resources according to your specific service requirement.
a. True
b. False
Answer:
a. True
Explanation:
Cloud computing can be defined as a type of computing that requires shared computing resources such as cloud storage (data storage), servers, computer power, and software over the internet rather than local servers and hard drives.
Generally, cloud computing offers individuals and businesses a fast, effective and efficient way of providing services.
In Computer science, one of the most essential characteristics or advantages of cloud computing is rapid elasticity.
By rapid elasticity, it simply means that cloud computing gives you the ability to expand and reduce resources according to your specific service requirement because resources such as servers can be used to execute a particular task and after completion, these resources can then be released or reduced.
Some of the examples of cloud computing are Google Slides, Google Drive, Dropbox, OneDrive etc.
Write an INSERT statement that adds this row to the Categories table:
CategoryName: Brass
Code the INSERT statement so SQL Server automatically generates the value for the CategoryID column.
Answer:
INSERT INTO categories (CategoryName)
VALUES ('Brass Code');
Explanation:
The SQL refers to the Structured Query Language in which the data is to be designed and maintained that occurred in the relational database management system i.e it is to be used for maintaining and query the database
Now the INSERT statement should be written as follows
INSERT INTO categories (CategoryName)
VALUES ('Brass Code');
Consider the following Stack operations:
push(d), push(h), pop(), push(f), push(s), pop(), pop(), push(m).
Assume the stack is initially empty, what is the sequence of popped values, and what is the final state of the stack? (Identify which end is the top of the stack.)
Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
