IB Maths AI HL · Paper 3

From $H$ the nearest is $A$ (40), then from $A$ the nearest is $E$ (50), from $E$ the nearest is $D$ (60), from $D$ the nearest is $C$ (90), then $B$ (80) and back to $H$ (85).

Start at $E$. Pick the shortest available edge each time: $EA$ (50), $ED$ (60), $DC$ (90), $CB$ (80).

Add the two shortest edges from $H$, which are $HA=40$ and $HE=70$.

The column for $D$ gives the probabilities of the next factory after visiting $D$, and $P(C)=0.4$.

The column for $C$ adds to 1: $0.1+0.2+0.2+q+0.2=1$.

The column for $D$ adds to 1: $0+0.2+0.4+r+0.2=1$.

Audrey starts at $A$.

Find the steady state vector on the calculator.

She travels from home to the factory and back, so each visit is $2\times$ the distance $H$ to the factory (A 40, B 85, C 160, D 100, E 70).

Multiply each number caught by its frequency and divide by 50.

Use the Poisson distribution with mean 2.88.

The expected frequency for 0 fish is about 2.8, which is less than 5, so the columns for 0 and 1 are combined to make the expected frequency big enough.

$H_0$: the number of fish caught per day can be modelled by a Poisson distribution.

There are 5 classes after combining, and the mean was estimated from the data.

The observed frequencies are $9,12,11,10,8$ and the expected frequencies are as given. Run the ${\chi }^2$ test with 3 degrees of freedom.

Since $0.843>0.05$, do not reject $H_0$. There is not enough evidence that the number of fish caught is not Poisson, so a Poisson model is reasonable.

$k$ must be positive. The length is increasing, so $\frac{dL}{dt}>0$, and $L<L_{\infty }$ means $L_{\infty }-L>0$. A positive product needs $k>0$.

The pollution could have affected $L_{\infty }$ (the fish may not grow as long, since lengths decreased) or $k$ (the fish may grow more slowly).

At $t=0$, $\frac{dL}{dt}=k(L_{\infty }-L_0)$. The graph is a positive, decreasing, concave-up curve (an exponential decay) starting at $\left(0,k(L_{\infty }-L_0)\right)$ on the vertical axis, with the horizontal asymptote $\frac{dL}{dt}=0$.