TimD1
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I have been unable to come up with anything resembling a formal proof, but I have noticed several things that may help someone do so. At each step, you must choose between moving a frog of color $A$ ...

$p^k$ represents the probability of exactly $k$ consecutive successful trials $q^{n-k}$ represents the probability of exactly $n-k$ consecutive failed trials However, we don't care about the order. ...

In this problem it's easier to first view it as three possible situations $\frac{3}{12}\times\frac{2}{11}$ Chance that they are both placed in first group $\frac{4}{12}\times\frac{3}{11}$ Chance ...

The easiest way to remember the formulas for converting polar to rectangular coordinates and vice versa is to draw the right triangle at the origin with sides $x$ and $y$, hypotenuse $r$, and angle $\... View answer 2 votes In short, you are correct. Here's a quick explanation of what each part of the equation physically represents, and how it would affect the graph of a cosine equation.$$H(t)= A + B \cos\left(Ct\right).... View answer 2 votes Yes, I think you're correct. Alternatively, once you have established that there are$2n-2$possible lines, you could argue that each line can either be drawn or not ($2$choices), and each set of ... View answer 2 votes First, note that in order to correctly solve "two consecutive tasks", you must solve the middle task, and then either the first or last (or both) as well. In other words, you must pass the middle task ... View answer 1 votes As you mention, the formula for removing left-recursion (and replacing it with right-recursion) involves replacing all instances of this rule: \begin{gather} A \rightarrow A\alpha\ |\ \beta\\ \end{... View answer Accepted answer 1 votes Here is a context-free grammar for the language you describe, assuming the alphabet is$\{0,1\}$: \begin{gather} S \rightarrow ST1\ |\ \epsilon\\ T \rightarrow 0\ |\ \epsilon\\ \end{gather} This ... View answer Accepted answer 1 votes Your reasoning in the question is correct, however your continued analysis in the comment is not. In this case, order is unimportant because the questions asks for the probability that the sum of the ... View answer 1 votes What the mean value theorem basically states is that on a continuous function, given two points$(x_1, y_1)$and$(x_2, y_2)$where$x_2 > x_1$, the function's derivative somewhere on the interval ... View answer 0 votes To solve the first question, use the equation for a line: \begin{gather*} y - y_1 = m(x - x_1) \end{gather*} For the second question, In order for$f(x)$to be continuous at$x = 8$, \begin{gather}... View answer 0 votes Estimate. Before solving a problem, quickly approximate what you would expect the ultimate solution to be. Before answering, ask yourself, it this value around what I initially expected? Does it make ... View answer Accepted answer 0 votes To construct a DFA which accepts all strings not containing 110, you can't just use the negation of a DFA which does accept 110. You must use the negation of the DFA which accepts exactly those ... View answer Accepted answer 0 votes Hint: Consider all possible placements of equilateral triangles (with side length 1000) placed on the globe. If none of these triangles have all three vertices in the water, then at least one out of ... View answer Accepted answer 0 votes The prime factors of$10$are$2$and$5$, so$10$can be rewritten as$2^15^1$. Since the prime factorization of any number is written as$2^a3^b5^c7^d11^e\dots$, the number of trailing zeros it has ... View answer Accepted answer 0 votes In normal division,$\frac{34}{5} = 6$remainder$4$. Or, you could re-express it by dividing the remainder by the original divisor as$6+\frac{4}{5}$. This corresponds directly to what is shown in ... View answer 0 votes You are correct. The probability that Player A wins$3$matches and loses$2$matches is$p^3q^2$, and there are${5 \choose 3}$ways in which this could happen. Therefore, the probability that Player ... View answer 0 votes The probability of escape is$\frac{2}{3}\$. There are six successful ways to escape: He chooses door 1 He chooses door 2, then 1 He chooses door 3, then 1 He chooses door 2, then 2, then 1 He ...