Recent content by romsek

Have you worked out the states and the state transition probability matrix? I see the states as Deuce, Ad-In, Ad-Out, Game with Game being an absorbing state.

No. $$(x,y) \to \left [0, \dfrac \pi 4\right]$$ is the same as $$x \to 0,~y \to \dfrac \pi 4$$ i.e the limit at the point $$\left(0,~ \dfrac \pi 4\right)$$ If you can plug them in and get a value, great, that's your limit. But generally plugging the values in will result in 0 in the...

what's the Hamming distance of the (7,4) code? (hint: it's 3) Can a valid codeword exist at a Hamming distance of 2 bits in a code with a Hamming distance of 3?

Newton's method solves this readily enough.

maybe they just need to move some goats! :ROFLMAO:

Matrix multiplication (note that a vector is just an m x 1 matrix) (n x m) X (m x k) => (n x k) See if you can figure it out now.

you can combine the numbers, $$-12+10-9$$ as well

You are omitting all the cases where two vowels are adjacent but the others are not, or where two pairs of vowels are adjacent but separated by a consonant.

2) is simple. Just look at how many valid digit choices you have for each digit of a 3 digit number. Note that 0 is not included as a possible digit and so we can't use the number 100 anyway so we don't have to worry that we don't use every 3 digit number. 3) Use stars and bars 4) Is a bit...

There are 2 sets of letter positions vowels can be placed in such that they are never adjacent. These are letters $$(1,3,5,7)$$, and letters $$(2,4,6,8)$$. Within these slots there would be $$4!=24$$ arrangements of the 4 vowels however there are 2 of the letter "A". Thus we have...

$$y^\prime = \dfrac{dy}{dx}$$ so just "multiply" both sides by $$dx$$ Have you read on how to solve separable diff eqs?

$$y^\prime = \dfrac{x^2}{y(1+x^3)}\\~\\ y dy = \dfrac{x^2}{1+x^3}dx\\~\\ \dfrac 1 2 y^2 = \dfrac{1}{3} \log \left(x^3+1\right) + C\\~\\ y^2 = \dfrac 2 3 \log(x^3+1) + C~\text{(note the constant absorbs multiplication by other constants)}\\~\\ y = \pm \left(\dfrac 2 3 \log(x^3+1) +...

It's a bit difficult to cite any approach as "best" given the variety of problems that exist but generally separating the cases helps ensure you won't multiply count specific situations.

With your method groups with 3 girls get counted 3 times each rather than once each.

We easily obtain that $$\dfrac{dy}{dx} = \dfrac{x(x^2+1)}{4y^3}\\ 4 y^3 dy = (x^3 + x) dx\\ y^4 = \dfrac 1 4 \left(x^4 + 2 x^2 + C\right)\\~\\ \text{It's easiest to evaluate the constant here}\\ (y(0))^4 = \dfrac 1 4 C = \left(-\dfrac{1}{\sqrt{2}}\right)^4 = \dfrac 1 4 \Rightarrow C = 1\\...