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If $c$ is the number of crackers, and $b$ is the number of birds, \left.\begin{align*} c&=5b+3\\ c&=6b-4 \end{align*}\right\}\implies5b+3=6b-4\implies 3+4=6b-5b\implies b=7

2

You forgot the integration constant. Following on from your integral for $T(t)$, we find \begin{align} T(t)e^{\alpha t} &= \int 2 \mu e^{\alpha t}\left(\frac{\theta}{\alpha} + \left(G -\frac{\theta}{\alpha}\right)e^{-\alpha t}\right)dt \\ &= \int 2 \mu e^{\alpha t}\left(\frac{\theta}{\alpha}\right) + 2 \mu \left(G -\frac{\theta}{\alpha}\right) dt \\ ...

1

Yes, the conditional probability that the coin will land with it's head up at $k$-th flip given the results of previous $k - 1$ flips equals simply the probability it will land with head up on the $k$-th flip, since outcomes of each flip are considered mutually independent (the coin has no memory, and outcomes of previous flips can in no way influence what ...

1

For any questions about coin tossing that involve an unbounded number of tosses the proper sample space $\Omega$ consists of all infinite binary sequences ${\bf b}=(b_1,b_2,\ldots)$. This space obtains a bona fide probability measure $\mu$ through a measure-theoretic construction starting from the requirement that the events $$A_n:=\{{\bf b}\in\Omega\>| ... 1 I am rewriting the problem as$$\left. y = \dfrac{x}{C_0+C_1 x+C_2 x^2} \right\}\;(C_0+C_1 x+C_2 x^2)y = x with the constants $C_0=C$, $C_1=-C D$ and $C_2=C D S^2$ and $x=M$, $y=R$ You want to find the coefficients $C_0$, $C_1$, and $C_2$ given 3 pairs of $(x_i,y_i)$ This can be treated as a 3×3 linear system when the three points are used in $C_0 y_i + ... 1 I'm not sure if this is what you want but perhaps you want to the codomain to be:$ \Delta(\Pi_{t\in T} F(x,y(t)))$or$\Pi_{t\in T}\Delta (F(x,y(t)))$The first is more general since it allows correlation across time. The second is more restrictive because assumes independence (and in this case we may have to assume$T\$ discrete as there are problems in ...

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