The image contains multiple questions related to a cyclist and their bike. The main question is "QUESTION 3". Specific sub-questions ask to calculate wheel radius, rotations, handlebar size, bottle volume, temperature conversion, and probability.

The diagram shows the height of the rear wheel as 70 cm, which represents its diameter.

$$\text{Diameter} = 70 \text{ cm}$$

The radius is half of the diameter.

$$\text{Radius} = \frac{\text{Diameter}}{2} = \frac{70 \text{ cm}}{2} = 35 \text{ cm}$$

The radius of the wheel is $\boxed{\text{35 cm}}$.

3.1.2

The competition distance is 23,7 kilometres.
We know that $1 \text{ km} = 1000 \text{ m}$ and $1 \text{ m} = 100 \text{ cm}$.
Therefore, $1 \text{ km} = 1000 \times 100 \text{ cm} = 100\,000 \text{ cm}$.

$$\text{Competition distance} = 23,7 \text{ km} \times 100\,000 \frac{\text{cm}}{\text{km}} = 2\,370\,000 \text{ cm}$$

The radius is 35 cm.

$$\text{Distance of one rotation} = 2 \times 3,142 \times \text{radius}$$ $$\text{Distance of one rotation} = 2 \times 3,142 \times 35 \text{ cm}$$ $$\text{Distance of one rotation} = 219,94 \text{ cm}$$

$$\text{Number of rotations} = \frac{\text{Total distance}}{\text{Distance of one rotation}}$$ $$\text{Number of rotations} = \frac{2\,370\,000 \text{ cm}}{219,94 \text{ cm}}$$ $$\text{Number of rotations} \approx 10\,775,666$$

The wheel would have to rotate approximately $\boxed{10\,775,67 \text{ times}}$.

3.1.3

The diagram shows an arrow spanning the entire length of the bike, labeled "Handle bar width 185 cm". The question states, "The size of the handlebar is roughly the distance between the two wheels." Given the diagram's label and the arrow indicating the overall length of the bike (from the front of the front wheel to the back of the rear wheel), it is most reasonable to interpret the 185 cm as the overall length of the bike, and thus, the approximate "size of the handlebar" as referred to in the question.

The size of the handlebar is $\boxed{\text{185 cm}}$.

3.1.4

The bottle has a diameter of 8 cm.

$$\text{Radius} = \frac{\text{Diameter}}{2} = \frac{8 \text{ cm}}{2} = 4 \text{ cm}$$

The height of the bottle is 22 cm.

$$\text{Volume} = 3,142 \times \text{radius}^2 \times \text{height}$$ $$\text{Volume} = 3,142 \times (4 \text{ cm})^2 \times 22 \text{ cm}$$ $$\text{Volume} = 3,142 \times 16 \text{ cm}^2 \times 22 \text{ cm}$$ $$\text{Volume} = 3,142 \times 352 \text{ cm}^3$$ $$\text{Volume} = 1105,984 \text{ cm}^3$$

We are given that $1 \text{ mL} = 1 \text{ cm}^3$.
Since $1 \text{ litre} = 1000 \text{ mL}$, then $1 \text{ litre} = 1000 \text{ cm}^3$.
The bottle's volume is $1105,984 \text{ cm}^3$.
Since $1105,984 \text{ cm}^3 > 1000 \text{ cm}^3$, the bottle is big enough.

The bottle's volume is $1105,984 \text{ cm}^3$, which is greater than 1 litre ($1000 \text{ cm}^3$). Therefore, the bottle $\boxed{\text{will be big enough}}$.

3.1.5

The formula is: $^\circ\text{F} = (1,8 \times ^\circ\text{C}) + 32^\circ$.
Given $^\circ\text{F} = 90^\circ$.

$$90 = (1,8 \times ^\circ\text{C}) + 32$$

Subtract 32 from both sides:

$$90 - 32 = 1,8 \times ^\circ\text{C}$$ $$58 = 1,8 \times ^\circ\text{C}$$

Divide by 1,8:

$$^\circ\text{C} = \frac{58}{1,8}$$ $$^\circ\text{C} \approx 32,22$$

The temperature in degrees Celsius is approximately $\boxed{32,22^\circ\text{C}}$.

3.1.6

The company sold seventeen bikes in total.

$$\text{Total number of bikes} = 17$$

There are four green bikes.

$$\text{Number of green bikes} = 4$$

$$\text{Probability (Green bike)} = \frac{\text{Number of green bikes}}{\text{Total number of bikes}}$$ $$\text{Probability (Green bike)} = \frac{4}{17}$$

The probability of selecting a green bike is $\boxed{\frac{4}{17}}$.