Question: 1. v = \sqrt{\frac{a}{b}} a = 6.43 \text{ correct to 2 decimal places} b = 5.514 \text{ correct to 3 decimal places} By considering bounds, work out the value of v to a suitable degree of accuracy. You must show all your working and give a reason for your final answer.

1. **State the problem:** We need to find the value of $$v = \sqrt{\frac{a}{b}}$$ where $$a = 6.43$$ (correct to 2 decimal places) and $$b = 5.514$$ (correct to 3 decimal places), by considering bounds to determine a suitable degree of accuracy. 2. **Determine the bounds for $$a$$ and $$b$$:** - Since $$a$$ is correct to 2 decimal places, its bounds are: $$6.425 \leq a < 6.435$$ - Since $$b$$ is correct to 3 decimal places, its bounds are: $$5.5135 \leq b < 5.5145$$ 3. **Calculate bounds for $$\frac{a}{b}$$:** - The minimum value of $$\frac{a}{b}$$ occurs when $$a$$ is minimum and $$b$$ is maximum: $$\frac{6.425}{5.5145} \approx 1.1655$$ - The maximum value of $$\frac{a}{b}$$ occurs when $$a$$ is maximum and $$b$$ is minimum: $$\frac{6.435}{5.5135} \approx 1.1675$$ 4. **Calculate bounds for $$v = \sqrt{\frac{a}{b}}$$:** - Minimum $$v$$: $$\sqrt{1.1655} \approx 1.0795$$ - Maximum $$v$$: $$\sqrt{1.1675} \approx 1.0806$$ 5. **Determine the suitable degree of accuracy:** - The bounds for $$v$$ are approximately $$1.0795 \leq v < 1.0806$$. - The difference between bounds is about $$0.0011$$. - This suggests that $$v$$ is accurate to 3 decimal places. 6. **Calculate the value of $$v$$ using the given values:** $$v = \sqrt{\frac{6.43}{5.514}} = \sqrt{1.1667} \approx 1.0801$$ 7. **Final answer:** $$v = 1.080$$ (to 3 decimal places) **Reason:** The bounds analysis shows that $$v$$ lies between approximately $$1.0795$$ and $$1.0806$$, so rounding to 3 decimal places is appropriate and consistent with the precision of the inputs.