Answer:
The five factors to consider when trying to choose between a Solid State Drive, a Hard Disk Drive and, an External Hard Disk Drive are:
- Read/Write Speed
- Weight
- Power Consumption
- Cost
- Storage Capacity
- Solid State Drives (SSDs) are typically lighter in weight, faster and do not consume much power.
- Hard Disk Drives are relatively cheaper than SSDs. They also come with higher storage capacities but are more power-hungry and slower because they rely on mechanical/moving parts to read and write data.
- External HDDs are the cheapest of the three. They are not internal which is a major drawback given the additional weight. However, they come with gargantuan storage capacities that make you want to rethink having one. Besides, unlike SSDs, you can easily get them in computer accessories shops offline or online.
Cheers!
The elements in a string type array will be initialized to "Null".
Answer:
#include <iostream>
using namespace std;
int main () {
// for loop execution
int semester_fees=8000;
int b=1;
cout<<"there are 5 years or 10 semester fees breakdown is given below"<<endl;
for( int a = 1; a <=5; a++ ) {
semester_fees = semester_fees*1.03;
cout<<"Semester fees for each of semester"<<b++<<"and"<<b++<<"is:$"<<semester_fees<<endl;
}
return 0;
}
Explanation:
code is in c++ language
Fees is incremented yearly and i have considered only two semester per year
Define variables
left is l
right is r
Ask input
left or right
Ask input value
Equate l or r to the input value
Show ladder with steps equal to input value and in the side of input variable
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
