Answer:
Jog
Explanation:
The variable options is a list containing 5 string values
options = ["ski", "surf", "jog", "bike", "hike"]
Indexing in python starts from 0 ; hence the index values of the list values are :
0 - ski
1 - surf
2 - jog
3 - bike
4 - hike
The statement ;
print(options[2]) means print the string at index 2 in the list named options
The string at index 2 is jog ;
Hence, the string jog is printed.
<span>The correct answer is higher for both blank spaces.
We all know the famous saying: "No risk, no reward". What is true is the higher your risk you also have a higher degree of reaping a higher rewards. But the opposite is also true, the more you risk the more you stand to lose. In stockbroker business this is best exemplified, as you can se brokers trying to predict the stock market in order to make greater profits. Gambling is also the good example of this. </span>
Answer : communicate is the answer.
Answer:
Hi Riahroo! This is a good question on the concept of relational databases.
We can normalize the relations as follows:
Flight
(flightnumber (unique), flighttime, airline_id, departure_city, arrival_city, passenger_id, pilot_id, airplane_id)
has_one_and_belongs_to :airline
has_many :passengers
has_one :pilot
Itinerary(passenger_id, flight_id)
Belongs_to
Passenger_details
(passengername (unique), gender, date_of_birth)
has_many :flights
Pilot
(pilotname (unique), gender, date_of_birth)
has_many :flights
airline(airlinename)
airplane(planeID, type, seats))
Explanation:
To normalize a relation, we have to remove any redundancies from the relationships between database objects/tables and simplify the structure. This also means simplifying many-to-many relationships. In this question, we see there is a many-to-many relationship between flights and passengers. To resolve this we can introduce a join table which simplifies this relationship to a one-to-many between the objects.
Answer:
// program in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// string array
string m[3];
// array to store rainfall
double rainfall[3];
// variables
double avg_rainfall,sum=0;
for(int i=0;i<3;i++)
{
cout<<"Enter name of month "<<i+1<<" :";
// read month name
cin>>m[i];
cout<<"Enter rainfall (inches) in month "<<i+1<<" :";
// read rainfall
cin>>rainfall[i];
// sum of rainfall
sum+=rainfall[i];
}
// Average rainfall
avg_rainfall=sum/3;
// print Average rainfall
cout<<"Average rainfall for "<<m[0]<<","<<m[1]<<","<<m[2]<<" is "<<avg_rainfall<<" inches."<<endl;
return 0;
}
Explanation:
Create string array "m" to store name of month and double array "rainfall" to store rainfall. Read name of 3 months and rainfall in that month.Find the sum of all the rainfall and the average rainfall.Print the average rainfall of 3 months.
Output:
Enter rainfall (inches) in month 2 :45
Enter name of month 3 :july
Enter rainfall (inches) in month 3 :43
Average rainfall for may,june,july is 42.6667 inches.
