$\xi = \{x : x \text{ is an integer},\ 2 \le x \le 14\}$ $A = \{x : x \text{ is a prime number}\}$ $B = \{x : x \text{ is a multiple of 3}\}$

1. Identify all elements in $\xi$ and in $A \cup B$.

$\xi=\{2,3,4,5,6,7,8,9,10,11,12,13,14\}$

Primes in $\xi$: $A=\{2,3,5,7,11,13\}$

Multiples of $3$ in $\xi$: $B=\{3,6,9,12\}$

So
$A\cup B=\{2,3,5,6,7,9,11,12,13\}$.

2. Find the complement of $A \cup B$ in $\xi$.

$(A\cup B)'$ means the elements in $\xi$ that are not in $A\cup B$.

$(A\cup B)'=\{4,8,10,14\}$

1. Find the element common to both $A$ and $B$.

In $\xi=\{2,3,4,5,6,7,8,9,10,11,12,13,14\}$,

$A=\{2,3,5,7,11,13\}$ and $B=\{3,6,9,12\}$.

The element common to both sets is $3$.

So,
$A\cap B=\{3\}.$

1. Identify possible elements of $C$ using the conditions.

Since $C \subset A$, the elements of $C$ must be prime numbers from $\xi$.

$A=\{2,3,5,7,11,13\}$.

Since $B \cap C=\emptyset$, $C$ must not contain any multiple of 3. The only element of $A$ that is also in $B$ is $3$, so exclude $3$.

Choose any 3 elements from $\{2,5,7,11,13\}$, for example

$C=\{2,5,7\}$.

1. Relate the empty small cone to the whole cone using the volume ratio of similar solids.

The liquid has half the volume of the cone, so the empty small cone also has half the volume of the cone.

The height of the empty small cone is $24-d$ cm, while the height of the whole cone is $24$ cm.

$\left(\frac{24-d}{24}\right)^3=\frac12$

2. Solve for the depth $d$.

$\frac{24-d}{24}=\sqrt[3]{\frac12}$

$24-d=24\sqrt[3]{\frac12}$

$d=24-24\sqrt[3]{\frac12}$

$d=4.952\ldots$

So, to 3 significant figures,

$d=4.95$

1. Relate the volume of cone T to cone P using the cone volume formula.

For a cone,
$V=\frac13\pi r^2h.$
Let cone P have radius $r$ and height $h$, so
$v=\frac13\pi r^2h.$

2. Apply the radius and height changes to find the volume of cone T in terms of $v$.

Cone T has radius $2r$ and height $\frac13h$.

$V_T=\frac13\pi(2r)^2\left(\frac13h\right)$
$=\frac13\pi\cdot 4r^2\cdot \frac13h$
$=\frac43\left(\frac13\pi r^2h\right)$
$=\frac43v.$