In a diagram RST, MN is parallel to ST. PM=5cm, RN=8cm and MS=10cm. Name the triangle similar to triangle PST, find the length of NT.

The mode is the value with the highest frequency. From the table, the highest frequency is 5, which corresponds to 3 children.
The mode is $\boxed{\text{3 children}}$.

The total number of houses is $\sum f = 15$. The position of the median is $\frac{N+1}{2}$, where $N$ is the total frequency.
Median position $= \frac{15+1}{2} = \frac{16}{2} = 8^{th}$ position.
Looking at the cumulative frequency, the 8th value falls in the category of 2 children (since the cumulative frequency reaches 5 at 1 child, and 9 at 2 children).
The median is $\boxed{\text{2 children}}$.

The mean is calculated as $\frac{\sum fx}{\sum f}$.
Mean $= \frac{30}{15} = 2$.
The mean is $\boxed{\text{2 children}}$.

: Geometry
Given the diagram with triangle PST and line segment MN parallel to ST.
Given: $PM = 5$ cm, $MS = 10$ cm.
There is a typo in the problem statement "RN = 8cm". Based on the diagram and typical similar triangle problems, we will assume 'RN' is a typo for 'PN', so $PN = 8$ cm.

Since MN is parallel to ST, by the property of similar triangles, $\triangle PMN$ is similar to $\triangle PST$.
The triangle similar to $\triangle PST$ is $\boxed{\triangle PMN}$.

First, calculate the length of PS:

Since $\triangle PMN \sim \triangle PST$, the ratio of corresponding sides is equal:

Substitute the known values:

Simplify the ratio:

Solve for PT:

Now, find NT:

The length of NT is $\boxed{\text{16 cm}}$.

: Vectors
Given: $\vec{AB} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ and $\vec{BC} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$.

Using the triangle law of vector addition, $\vec{AC} = \vec{AB} + \vec{BC}$.

The column vector for $\vec{AC}$ is $\boxed{\begin{pmatrix} 9 \\ 0 \end{pmatrix}}$.