To solve part (ii), we need to find the equation of the straight line graph plotted in (a) and hence find the values of $a$ and $b$.

The relationship between $y$ and $x$ is given by $y = ab^x$.

Given data:

• $x = 15, 20, 25, 30, 35$
• $y = 2.4, 4.4, 8.0, 15.5, 28.6$

Steps:

1. Linearize the equation: Take the logarithm of both sides of $y = ab^x$: $\log_{10}(y) = \log_{10}(a) + x\log_{10}(b)$

   This equation is in the form $\log_{10}(y) = mx + c$ where:

   • $m = \log_{10}(b)$
   • $c = \log_{10}(a)$

2. Calculate $\log_{10}(y)$ values: Compute the $\log_{10}(y)$ for each $y$.

   • $\log_{10}(2.4) \approx 0.38$
   • $\log_{10}(4.4) \approx 0.64$
   • $\log_{10}(8.0) \approx 0.90$
   • $\log_{10}(15.5) \approx 1.19$
   • $\log_{10}(28.6) \approx 1.46$

3. Plot and find the equation of the line: Using the graph from part (a), plot $(x, \log_{10}(y))$ points and draw the best-fit line.

4. Determine values from the equation $\log_{10}(y) = mx + c$:

   • Find the slope $m$ and intercept $c$ from the graph.
   • $m \approx 0.057$, $c \approx -0.48$ (use your graph for precise values).

5. Calculate $a$ and $b$:

   • $\log_{10}(b) = m \Rightarrow b = 10^m \approx 10^{0.057} \approx 1.14$
   • $\log_{10}(a) = c \Rightarrow a = 10^c \approx 10^{-0.48} \approx 0.33$

Final Answer:

• $a \approx 0.33$
• $b \approx 1.14$

Round to one significant figure:

• $a \approx 0.3$
• $b \approx 1.1$