IB Maths AI HL · Paper 3

The cycle $A\to B\to C\to D\to E\to F\to A$ uses the edges of weights $8,7,9,6,12,5$.

The shortest path from $B$ to $D$ is $B\to A\to D$ (the cable car), which takes $8+4$.

The shortest path from $C$ to $F$ is $C\to A\to F$, which takes $12+5$.

Start at $A$ and move to the nearest unvisited vertex each time, using Table 1: $A\to D$ (4), then $D\to E$ (6), then $E\to F$ (11), then $F\to B$ (13), then $B\to C$ (7), then back to $A$ (12).

Prim's algorithm on Table 2 starting from $B$: take the edge $BC$ (7), then $CD$ (9), then $DE$ (6), then $DF$ (9).

Add back the two shortest edges at $A$ from Table 1, which are $AD=4$ and $AF=5$.

The largest lower bound is 40, so $p=40$.

The smallest upper bound is the smaller of 47 and 53.

The route is $A\to E\to D\to C\to B\to F\to A$. The walks are $AE=6$, $ED=6$, $DC=9$, $CB=7$, and from $B$ to $F$ the shortest way is through $A$ ($8+5=13$), and then $FA=5$.

The way from $B$ to $F$ goes through $A$ and uses the edge $AF$, which is also used for the last leg, so

Add the means and add the variances.

The time available is 150 minutes. Walking takes 46 minutes and the five rides take $5\times 2=10$ minutes, so queueing must take less than $150-46-10=94$ minutes.

The assumption is that 200 is the largest population the area can support (the carrying capacity), because the historical stable size was 200.

Put $t=0$ and $x=40$.

Put $t=5$ and $x=70$.

At $t=10$.

When $x=200$, $\ln \frac{200}{200}=\ln 1=0$.

At 200 wolves the population is not changing, so 200 is the steady (equilibrium) population.

Differentiate $f(x)=\ln \left(\ln 200-\ln x\right)$ with the chain rule.

Use $x=40$ when $t=0$.

Use $x=70$ when $t=5$.

At $t=10$.

After 10 years the real population is 85. The logistic model gives 107 and the Gompertz model gives 101, so both overestimate. The Gompertz prediction is closer.

The estimate of 89 is much closer to the measured 85 than either earlier model, so allowing for emigration makes the model more realistic.