IB Maths AI HL · Paper 3

1. Order the rows and columns $A,B,C,D,E$ and put a $1$ wherever two junctions share a road, $0$ otherwise.
2. No road loops back to its own junction, so the leading diagonal is all zeros, and the matrix is symmetric because the roads are two-way.

1. The number of walks of length 4 between two vertices is the matching entry of ${\mathbf{M}}^4$.
2. Raise the matrix to the fourth power on the GDC and read row $A$, column $D$.

1. Using Kruskal: list the roads from cheapest to dearest and add each one unless it closes a cycle.
2. Stop once all five junctions are joined, which takes $5-1=4$ roads.

Another way. Prim's from $A$ picks $AD$, then $BD$, $CD$ and $DE$: the same tree.

1. Every road must be swept and he returns to $A$, so this is a Chinese postman problem: we need a closed walk covering every edge.
2. Degrees: $A=3$, $B=4$, $C=2$, $D=4$, $E=3$. Only $A$ and $E$ are odd.
3. The two odd vertices must be joined by repeating the shortest path between them. Direct $AE$ takes $9$; via $D$ it takes $4+6=10$. So repeat $AE$.
4. Minimum time is the total of all roads plus the repeated $9$.

1. With $AE$ at $18$, sweeping it once adds $9$ to the base total.
2. The repeat between $A$ and $E$ should now go via $D$ ($4+6=10$) rather than along $AE$ again ($18$).
3. Compare with the original $55$. Any other road doubling adds at most $8$ (road $BE$), so this is the worst case.

1. No stops means no waiting, so only one or two stops can give at least $4$ minutes.
2. One stop: the wait is $W\sim N(3,0.8^2)$. Find $P(W\ge 4)$.
3. Two stops: the total of two independent waits is normal with mean $3+3=6$ and variance $0.64+0.64=1.28$. Find $P(W_1+W_2\ge 4)$.
4. Weight each by the chance of that many stops from (e)(ii) and add.

1. The data only cover concentrations from $1.0$ to $4.0$ mol, and $10.0$ mol is well outside that range.
2. Using the line there is extrapolation, so the estimate is not reliable and not appropriate, however strong $r$ is.

1. Differentiate $bC+aD$ with respect to $t$ and substitute both rates from the model.
2. The two terms cancel, so the derivative is zero for all $t$, which means $bC+aD$ is constant.

Another way. Divide the two equations to get $\frac{\mathrm{d}C}{\mathrm{d}D}=-\frac{a}{b}$, integrate to $C=-\frac{a}{b}D+k$ and rearrange.

1. Euler's method: $D_{n+1}=D_n+h\cdot f(D_n)$, with $f(D)=0.025D(3.08-0.8D)$, $h=5$ and $D_0=0.1$.
2. Going from $t=0$ to $t=80$ takes $16$ steps. Run the recursion on the GDC (a sequence or spreadsheet mode is quickest).
3. Record a few early values so the method is visible, then give $D$ at $t=80$.

1. Enter the summary statistics into the GDC's two-sample $t$-test with pooling switched on. Use the standard deviations $s_{n-1}=\sqrt{0.0784\dots }$ and $\sqrt{0.0440\dots }$ with $n=15$ each.
2. Compare the $p$-value with $0.05$ and write the conclusion in context.

Another way. Stating that the two samples are independent of each other is also a valid assumption.