Answer:
if(revenue.cents - expenses.cents < 0){
profit.dollars = revenue.dollars - expenses.dollars - 1;
profit.cents = 1 - revenue.cents - expenses.cents;
}
else{
profit.dollars = revenue.dollars - expenses.dollars;
profit.cents = revenue.cents - expenses.cents;
}
Explanation:
We know that profit is given as: revenue - expenses from the question.
From the given expression above;
if(revenue.cents - expenses.cents < 0)
then profit.dollar will be revenue.dollars - expenses.dollars - 1; the 1 is to be carry over to the cent part. And the profit.cent will be 1 - revenue.cents - expenses.cents;
else the profit.dollars and the profit.cent is computed directly without needing to carry over:
profit.dollars = revenue.dollars - expenses.dollars;
profit.cents = revenue.cents - expenses.cents;
Answer:
The correct answer is B. Africa
Explanation:
African Institute for Mathematical Sciences (AIMS) was founded in 2003 having its first center in Cape Town, South Africa. AIMS provides training for Africa's talented university graduates needed to enter technical professions or pursue graduate studies in technical fields. The Next Einstein Initiative (NEI) is a strategic plan to build on the success of the first AIMS centre and create a coordinated pan-African network of 15 AIMS centres by 2020, producing 750 well-qualified graduates per annum. AIMS won the Google's Project 10^100 and also the founder of AIMS won the TED Prize and announced his vision to unlock scientific talent across Africa.
The long term memory used by the computer is called “RAM”
Answer:
Check the explanation
Explanation:
#!usr/bin/python
#FileName: sieve_once_again.py
#Python Version: 2.6.2
#Author: Rahul Raj
#Sat May 15 11:41:21 2010 IST
fi=0 #flag index for scaling with big numbers..
n=input('Prime Number(>2) Upto:')
s=range(3,n,2)
def next_non_zero():
"To find the first non zero element of the list s"
global fi,s
while True:
if s[fi]:return s[fi]
fi+=1
def sieve():
primelist=[2]
limit=(s[-1]-3)/2
largest=s[-1]
while True:
m=next_non_zero()
fi=s.index(m)
if m**2>largest:
primelist+=[prime for prime in s if prime] #appending rest of the non zero numbers
break
ind=(m*(m-1)/2)+s.index(m)
primelist.append(m)
while ind<=limit:
s[ind]=0
ind+=m
s[s.index(m)]=0
#print primelist
print 'Number of Primes upto %d: %d'%(n,len(primelist))
if __name__=='__main__':
sieve()
