Answer:
Overall project duration
Explanation:
Scheduling can best be defined as the process used to determine a overall project duration.
Answer:
a) 
b) 
Explanation:
Previous concepts
The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".
Part a
Let X the random variable of interest. We know on this case that 
And we know the probability denisty function for x given by:

In order to find the cdf we need to do the following integral:

Part b
Assuming that
, then the density function is given by:

And for this case we want this probability:

And evaluating the integral we got:

Answer:A. No one has ever beat Nancy.
Explanation:
The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.
The dormain of discourse is also known simply as universe, can also be said to be a set of entities o
upon which certain variables of interest in some formal treatment may range.
The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.
THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.
Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness,
is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for
while a is taken as 0.003m and Y is already known

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material