Answer:
$$\begin{align*}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Explanation:
\eqalign{
P(Y-X=m\mid Y\gt X)
&=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr
&=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr
&=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr
}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Answer:
road closures on deliver routes.. i think..
Answer:
a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s
Explanation:
Diffusion is governed by Arrhenius equation
I will be using R in the equation instead of k_b as the problem asks for molar activation energy
I will be using
and
°C + 273 = K
here, adjust your precision as neccessary
Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm
So:
and
You might notice that these equations have the form of
You can solve this equation system easily using calculator, and you will eventually get
After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation
And you should get D = 2.76*10^-16 m^/s as an answer for c)