IB Maths AI SL · Paper 2

Gradient through $(1708,8.73)$ and $(1958,9.45)$.

The gradient is the rate at which the mean annual temperature rises: it increases by 0.00288 degrees each year.

Use the point $(1708,8.73)$.

Substitute $x=2000$.

Use linear regression of $y$ on $x$ on the seven points.

Substitute $x=2000$.

The term $\frac{4}{x^2}$ is not defined when $x=0$.

Find the minimum on the right-hand branch with the calculator, or solve $f^{\prime }(x)=3-\frac{8}{x^3}=0$.

The right branch has a minimum value 5.24. A horizontal line with $w$ above 5.24 meets the right branch twice and the left branch once.

Use the two points $A(0,-5)$ and $B(3,4.5)$ in $g(x)=k{p}^x-9$.

As $x\to -\infty$ the term $p^x\to 0$ because $p>1$.

Solve $f(x)=g(x)$ for $x>0$ with a graph.

The amplitude is half the difference between high and low depth.

The centre line is the mean of the high and low depths.

From 06:00 to 12:15 is $6\times 60+15=375$ minutes, half a period.

The period is $\frac{2\pi }{b}$.

The depth is below 6 m from $t=282.5$ until the symmetrical time $750-282.5=467.5$, which lies before 15:00 ($t=540$).

From the user branch (0.06): positive 0.71 and negative 0.29. From the non-user branch (0.94): positive 0.02 and negative 0.98.

Each Team X athlete tests negative with probability 0.98, independently.

The number of positive tests is $X\sim B(20,0.02)$.

The container is half of a cylinder, so $V=\frac{1}{2}\pi r^{2}l=0.8$.

Differentiate term by term.

At the minimum $\frac{dC}{dr}=0$ with $r=0.7$.

The total before discount is $8641, which is in the band $5000\le C<10000$, so the discount is 8%.