IB Maths AI SL · Paper 2

(i) Tablet row total over the grand total.

(ii) The cell for 11–13 and mobile phone.

(iii) Given the student is 17–18, the sample space is the 348 students in that age group.

(iv) Add the tablet total and the 14–16 total, and subtract the overlap (224).

State the hypotheses about association.

For a $3\times 3$ table the degrees of freedom are

(i) The calculator gives the test statistic.

(ii) The $p$-value.

(iii) $23.3>13.277$ (equivalently $p<0.01$), so reject $H_0$. There is sufficient evidence that preferred device and age group are not independent.

Percentage increase is the change over the original.

(i) The common ratio is 1.03.

(ii) The first term is 12 300.

(iii) Put in $n=11$.

The places form an arithmetic sequence with first term 10 380 and difference 600.

The total number of places in 10 years is the sum of the first 10 terms, and each pays $80.

Not for all $n>k$. Applications grow exponentially and places grow linearly, so applications will eventually overtake places again (from about $n=24$).

The tower is 90 m high and the blade is 40 m long.

(i) 12 rotations a minute means one every 5 seconds.

(ii) $360^{\circ }$ in 5 seconds.

In $h(t)=90-40\cos (72t^{\circ })$:

Start at the minimum, $(0,50)$. The graph rises to the maximum $(2.5,130)$ and falls back to a minimum at $(5,50)$.

(i) At $t=2$:

(ii) Solve $90-40\cos (72t)=100$, which gives $t\approx 1.451$ and $3.549$ in one rotation.

(i) Tim sees C only when it is at most 100 m high. In each 5-second rotation C is above 100 m for 2.098 seconds, so it is visible for the rest.

(ii) The blades turn at twice the speed, so each rotation takes half as long, but C still spends the same fraction of each rotation below 100 m.

The bottom is 10 m in from the edge on each side, over a depth of 6 m.

(i) By similar triangles, or using $\tan \theta =0.6$ with the half-width 40:

(ii) The reservoir is the large pyramid with apex V minus the small pyramid below it, whose height is $24-6=18$.

With $X\sim \mathrm{N}(10,3^2)$:

The throws are independent, so multiply.

The chance of a point on a turn is $0.8143^3$.