Answer:
C. bachelor’s degree in filmmaking
Explanation:
Because Andy wants to become a multimedia producer, the degree that would best help him achieve his goal is a bachelor's degree in filmmaking.
Multimedia has to do with both audio, video and graphics or animations because it encompasses multiple media files.
Filmmaking has to do with the various forms of making and producing films. Filmmaking has to do with multiple media (multimedia) and it would help him achieve his goal.
Answer:
B decelerate
Explanation:
Deceleration will help in avoiding you from loosing control of the vehicle
Solution :
class Employee:
#Define the
#constructor.
def __
__(
, ID_number, 
, email):
#Set the values of
#the data members of the class.
= name
_number = ID_number
= salary
self.email_address = email
#Define the function
#make_employee_dict().
def make_employee_dict(list_names, list_ID, list_salary, list_email):
#Define the dictionary
#to store the results.
employee_dict = {}
#Store the length
#of the list.
list_len = len(list_ID)
#Run the loop to
#traverse the list.
for i in range(list_len):
#Access the lists to
#get the required details.
name = list_names[i]
id_num = list_ID[i]
salary = list_salary[i]
email = list_email[i]
#Define the employee
#object and store
#it in the dictionary.
employee_dict[id_num] = Employee(name, id_num, salary, email)
#Return the
#resultant dictionary.
return employee_dict
Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
Answer:
def words_in_both(a, b):
a1 = set(a.lower().split())
b1 = set(b.lower().split())
return a1.intersection(b1)
common_words = words_in_both("She is a jack of all trades", 'Jack was tallest of all')
print(common_words)
Explanation:
Output:
{'all', 'of', 'jack'}
