hmm wrong forum maybe this should go to programming and computer science instead? Can I get a mod to help me out
edit: actually please delete this thread I have my answer now :) thank you
In bash, I am writing the following:
ps - ly
this command gives me output of three lines
then I do:
ps -ly | wc -l
to read the number of lines. It says 4.
I clearly see that there are three lines. Thinking that there might be a new line possibly, I open export the command output to a file...
I've read quantum quests data structures bible but I'm still confused. I need help in simplest terms understanding what an ADT actually is and how it's different from a data structure.
So a data structure is a way of organizing data. I can understand why a stack, or a queue would fall under...
That is one amazing resource man. I am only able to skim through it right now but I can already tell I'm going to be coming back to that a LOT! Much love for that man
Problem Statement: I've attached a picture of his notes. For a sorted array, to update it he put: O(logn)+O(1)
Relevant Equations: O(n) + O(logn) = O(n)
I don't see how it can be O(logn)+O(1) and not O(logn)+O(n) in the worst case
After you find the index that needs updating, and you...
The answer says insertion sort runs faster when we're sorting less than 43 items. I agree but with the condition that the first item will not be faster. Why does the answer not mention this? Is it because it is insignificantly faster?
+(3/2) standard deviations from the mean = \frac {a+b}{12} + \frac{\sqrt3}{4} (b-a)
-(3/2) standard deviations from the mean = \frac {a+b}{12} - \frac{\sqrt3}{4} (b-a)
\frac {1}{b-a} \int_a^{\frac {a+b}{12} - \frac{\sqrt3}{4} (b-a)} dx = m_1= \frac {(-11+3\sqrt3)a + (1-3\sqrt3)b}{12(b-a)}...
I answered this question:"
b)Given that the system failed during the first 2 years, what is the probability that it
failed due to the failure of component B (but not component A)? "
Using:
P{before 2 years A doesnt fail but B fails} = .6147
divided by:
P{system does fail before 2 years} =...
My book never mentions Erlang. I understand the third paragraph.
What I understand about Gamma distributions:
When a procedure consists of α amount of steps and each step takes an exponential amount of time λ, then the total time has a gamma distribution.
What I think I did wrong in this...
Sorry most of my threads from this course have been in this forum so I thought it would fit.
And yes, I do mean to use 'alpha' but I had trouble finding it :/
I'm lost. First one was easy to calculate, second one is harder.
I have:
P{a fails before 2 yrs} = .323325
P{b fails before 2 yrds} = .90844
P{system doesnt fail for 2 years or longer} = .062
P{system does fail before 2 years} = .938
P{A and B fail before 2 yrs} = .29372
P{before 2 years A...
there must be an even number of vertices of odd degree, and from the handshake theorem, 2m = 2(12) = 24
the only way we can get this from 6 vertices using 2 and 5 is:
4 vertices of degree 5, 2 vertices of degree 2
does this seem correct??
oh wasn't a hw question, was just wondering myself. I appreciate it though! Also, I use to be able to mark my thread as solved with the old layout. Where is the button on this new one??
I would assume that it has some area even if it is really really small. But I guess a line implies that the left and right boundaries are going to the middle an infinite amount, so it has area =0? does anyone get what I mean?
Nice! this problem is actually from a discrete math book, and I'm taking a probability course right now, and we haven't learned about hypergeometric distribution surprisingly.
I don't understand why I can't answer this question as a bernuli trial.
There are 6 possible correct integers out of 40, and 34 incorrect integers out of 40. I'd assume it would look like this:
(6c1)(6/40)(34/40)^5
I guess, it's because when you choose and incorrect or correct integer, the...
Ahh sorry, meant to write:
x-y:
\begin{array}{|c|c|c|c|}
\hline x-y & -1 & 0 & 1 \\
\hline p(x-y) & .2 & .6 & .2 \\
\hline
\end{array}
The probabilities add up to 1 so I think its correct? Does this seem right to you?
Hmm I think what I was doing wrong is computing X+Y and X-Y from the marginal distribution, when I should be computing it from the joint distribution
so for
:x+y
\begin{array}{|c|c|c|c|}
\hline x+y & 0 & 1 & 2 \\
\hline p(x+y) & .5 & .4 & .1 \\
\hline
\end{array}
not
for x+y...