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Using gross margin to calculate selling price Given the cost of an item, one can compute the selling price required to achieve a specific gross margin. For example, if your product costs $100 and the required gross margin is 40%, then Selling price = $ 100 1 − 40 % = $ 100 0.6 = $ 166.67 {\displaystyle {\text{Selling price}}={\frac {\$100}{1 ...
In chemistry, the mass fraction of a substance within a mixture is the ratio (alternatively denoted ) of the mass of that substance to the total mass of the mixture. [1] Expressed as a formula, the mass fraction is:
In chemistry, the lever rule is a formula used to determine the mole fraction (x i) or the mass fraction (w i) of each phase of a binary equilibrium phase diagram.It can be used to determine the fraction of liquid and solid phases for a given binary composition and temperature that is between the liquidus and solidus line.
As an example, a sample with 70 % of R isomer and 30 % of S will have a percent enantiomeric excess of 40. This can also be thought of as a mixture of 40 % pure R with 60 % of a racemic mixture (which contributes half 30 % R and the other half 30 % S to the overall composition).
The COGS formula is the same across most industries, but what is included in each of the elements can vary for each. It should be calculated as: Operating Profit Margin = 100 ⋅ Operating Income Revenue {\displaystyle {\text{Operating Profit Margin}}={100\cdot {\text{Operating Income}} \over {\text{Revenue}}}}
In atmospheric chemistry, mixing ratio usually refers to the mole ratio r i, which is defined as the amount of a constituent n i divided by the total amount of all other constituents in a mixture: r i = n i n t o t − n i {\displaystyle r_{i}={\frac {n_{i}}{n_{\mathrm {tot} }-n_{i}}}}
The volume of such a mixture is slightly less than the sum of the volumes of the components. Thus, by the above definition, the term "40% alcohol by volume" refers to a mixture of 40 volume units of ethanol with enough water to make a final volume of 100 units, rather than a mixture of 40 units of ethanol with 60 units of water.
An aqueous solution containing 2 g of glucose and 2 g of fructose per 100 g of solution contains 2/100=2% glucose on a wet basis, but 2/4=50% glucose on a dry basis.If the solution had contained 2 g of glucose and 3 g of fructose, it would still have contained 2% glucose on a wet basis, but only 2/5=40% glucose on a dry basis.