A solid sphere of radius 3 cm is placed inside a cylinder and then water is poured into the cylinder until it is full, as shown in the diagram - Leaving Cert Mathematics - Question 8 - 2019

To find the drop in height, we will first calculate the volume of water displaced by the sphere:

The volume of the sphere is already calculated as $36\pi \, \text{cm}^3$. The volume of water in the cylinder can be calculated using the formula:

$V = \pi r^2 h$

Letting $r = 5$ cm (the internal radius of the cylinder), we can solve for the height when the volume is equal to the volume displaced by the sphere:

$36\pi = \pi (5)^2 h$

Dividing both sides by $\pi$:

$36 = 25h$

Therefore: $h = \frac{36}{25} = 1.44 \, \text{cm}$

The drop in height of the water is 1.44 cm.

First, we calculate the curved surface area (C.S.A.) of the cylinder using the formula:

$C.S.A. = 2\pi rh$

Substituting $r = 5\text{ cm}$ and $h = 18 \text{ cm}$:

$C.S.A. = 2\pi (5)(18) = 180\pi \, \text{cm}^2$

Next, we find the area of the rectangular piece of metal:

$\text{Area} = 35 \text{ cm} \times 20 \text{ cm} = 700 \, \text{cm}^2$

To get the area of the metal left over after cutting out the cylindrical part:

$\text{Area left} = 700 - (180\pi)$

Calculating $180\pi$ (using $\pi \approx 3.14$):

$180\pi \approx 565.49 \, \text{cm}^2$

So, $\text{Area left} = 700 - 565.49 \approx 134.51 \, \text{cm}^2$

Rounded to one decimal place, the area left is $134.5 \, \text{cm}^2$.

The margin of error (ME) can be calculated using the formula:

$ME = \frac{1}{\sqrt{n}}$

where $n = 800$. Thus: $ME = \frac{1}{\sqrt{800}} \approx 0.0355$

To express this as a percentage: $ME \approx 3.55\%$

The sample proportion is: $\hat{p} = \frac{350}{800} = 0.4375$

The 95% confidence interval can be calculated using: $\hat{p} \pm ME$

Thus, substituting the margin of error: $0.4375 \pm 0.0355$

The confidence interval is approximately: $[0.4375 - 0.0355, 0.4375 + 0.0355]$ $[0.4020, 0.4730]$

Expressed as percentages: $[40.20\%, 47.30\%]$

We will conduct a hypothesis test with null hypothesis: $H_0: \text{The level of support is 50\%}$

And alternative hypothesis: $H_1: \text{The level of support is not 50\%}$

The 95% confidence interval found in (b)(ii) was $[40.20\%, 47.30\%]$.

Since 50% falls outside this interval, we reject the null hypothesis. Therefore, the evidence suggests that the level of support for the proposal is not 50%.

Conclusion:

The support is significantly different from 50%.

Reason:

The confidence interval does not contain 50%, indicating that the level of support is likely less than 50%.