A baker uses (3)/(4) kg of flour, (5)/(6) kg of sugar, and (1)/(3) kg of cocoa. He sells the mixture for GHC 120 per kg. Find the total mass of the mixture and the total sales.

Mass of flour $= \frac{3}{4}$ kg
Mass of sugar $= \frac{5}{6}$ kg
Mass of cocoa $= \frac{1}{3}$ kg
Total mixture $= \frac{3}{4} + \frac{5}{6} + \frac{1}{3}$ kg

The LCM of 4, 6, and 3 is 12.

Total mixture $= \frac{9}{12} + \frac{10}{12} + \frac{4}{12} = \frac{9 + 10 + 4}{12} = \frac{23}{12}$ kg

The total mixture is $\boxed{\frac{23}{12} \text{ kg}}$.

c)

Selling price per kg = GHC 120

Total sales $= \text{Total mixture} \times \text{Price per kg}$

The total sales are $\boxed{\text{GHC } 230}$.

2. a)

To move the decimal point after the first non-zero digit (6), we move it 4 places to the right.

The number 6.84 already has 3 significant figures.
So, 0.000684 in standard form and rounded to 3 significant figures is $\boxed{6.84 \times 10^{-4}}$.

b)

Divide both sides by 8.

The solution is $\boxed{x=3}$.