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Answer: b)because near-critical paths could become critical paths with small delays in these activities
Explanation: Program evaluation and review technique(PERT) technique is used for management of the project .The organizing, maintenance,coordination of any task is done by this chart. The non-critical activities need to be observed carefully because they may create near-critical methods which can lead to complexity.
Other options are incorrect because not all activities are equally rather monitoring is done to avoid errors and slack can occur at any path . Small delays can be caused but not complete project can get delayed and there can be risk of complexity rather than being uncompleted.Thus, the correct option is option(b).
Answer:
There is no need to make an algorithm for this simple problem. Just add the two numbers by storing in two different variables as follows:
Let a,b be two numbers.
c=a+b;
print(c);
But, if you want to find the sum of more numbers, you can use any loop like for, while or do-while as follows:
Let a be the variable where the input numbers are stored.
while(f==1)
{
printf(“Enter number”);
scanf(“Take number into the variable a”);
sum=sum+a;
printf(“Do you want to enter more numbers? 1 for yes, 0 for no”);
scanf(“Take the input into the variable f”);
}
print(Sum)
Explanation:
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Answer:
import pandas as pd #importing pandas library as pd
import matplotlib.pyplot as plt #importing matplotlib.pyplot as plt
pop=pd.read_csv('nycHistPop.csv') #reading the csv file
borough=input('Enter borough name:') #asking the user for borough namme
# image=input('Enter image name:')
# pop['Fraction']=pop[borough]/pop['Total']
# pop.plot(x='Year', y='Fraction')
print("Minimum population",pop[borough].min()) #printing the minimum population of borough
print("Maximum population",pop[borough].max()) #printing the maximum population of borough
print("Average population",pop[borough].mean()) #printing the average population of borough
print("Standard deviation",pop[borough].std()) #printing the standard deviation of borough
# fig=plt.gcf()
# fig.savefig(image)
Explanation:
As specified in RFC5735, this is an address from the "link local" block. It is assigned to a network interface as a temporary address, for instance if no static address is configured and the DHCP server is not found.
If you boot your PC without a network cable, you'll probably end up with a 169.254.*.* address.
