IB Maths AI SL · Paper 2

Read the box plot: the median is at 4, the upper quartile at 6 and the lower quartile at 3.

Use the mid-interval values $1.5,3.5,5.5,7.5,9.5,11.5$ with the frequencies. The total frequency is 1260.

The alternative hypothesis.

(i) There are 7 classes.

(ii) The statistic, from the calculator or from the formula.

(iii) The $p$-value for ${\chi }^2=20.3$ with 6 degrees of freedom.

Since $20.3>16.812$ (equivalently $0.00244<0.01$), reject $H_0$. There is sufficient evidence at the 1% level that the word lengths are not consistent with the English language.

(i) The rate is $\frac{dh}{dt}=1.2-0.1t$. At $t=3$:

(ii) Three days after planting, the plant is growing at 0.9 cm per day.

The maximum height happens when the rate is zero.

The maximum height is at $t=12$.

Put $t=0$.

The multiplier each day is 1.2, which is an increase of 20%.

(i) The flying students are $21+14=35$ of the 50.

(ii) Of the 35 flying students, 14 also take management.

$F$ and $M$ are not mutually exclusive, because 14 students are enrolled in both courses, so $P(F\cap M)\ne 0$.

(i) The management students are $14+x$, and $P(F\mid M)=\frac{14}{14+x}=0.7$.

(ii) All the students are 50, so $y$ is what is left outside the two circles.

(iii) The students in only one course: flying only, management only, and engineering (outside the circles).

Choose three of the 35 flying students without replacement.

The graph passes through $P(0,4)$, so $c=4$.

(i) Point $R(5,3.9)$ is on the curve.

(ii) From point $Q$: $4a+2b=-0.064$. Substitute $b=-0.02-5a$.

(i) Differentiate.

(ii) The gradient has its greatest size at the ends of the bridge, $x=0$ or $x=10$.

(i) Use the calculator.

(ii) This is very close to 1, so there is a very strong positive correlation between the number of people and the deflection.

The gradient 0.01889 means the deflection increases by about 0.01889 m for each extra person on the bridge.

The centre is at $3+25=28$ m. $V$ is four chairs round from $M$ (counting anticlockwise from the lowest point), which is $4\times 30^{\circ }=120^{\circ }$ from the bottom, so it is $30^{\circ }$ above the horizontal through the centre.

After 9 minutes the wheel has turned $\frac{9}{12}=\frac{3}{4}$ of a revolution, so $M$ is level with the centre.

The amplitude of $h(t)=-25\cos bt+d$ is the radius.

(i) One revolution takes 12 minutes, and in degrees $b=\frac{360}{12}$.

(ii) The midline is the height of the centre.

The beam turns about $C$. The point $B$ is 20 m from $C$ (because its height varies by 20 m on each side of 40 m), and $K$ is 10 m from $B$. The furthest $K$ can be from $C$ is when it is in line with $B$ and $C$.