The methods described to be used to analysis of ore that assay about 1.2% of gold what minimum sample mass should be taken if relative error resulting from 0.4%mg loss is not exceed a) -0.1& -0.8%?

To determine the minimum sample mass that should be taken, we need to consider the relative error resulting from a 0.4% mass loss.

Let's assume the initial mass of the ore sample is M grams.

The mass of gold in the ore sample is 1.2% of M grams, which is 0.012M grams.

If there is a 0.4% mass loss, the final mass of the gold in the sample would be (1 - 0.004) * 0.012M grams.

The relative error can be calculated as the difference between the initial and final mass of gold, divided by the initial mass of gold:

Relative error = [(1 - 0.004) * 0.012M - 0.012M] / 0.012M

Simplifying the equation:

Relative error = (0.012M - 0.0048M - 0.012M) / 0.012M

Relative error = -0.0048M / 0.012M

Relative error = -0.4

To ensure that the relative error does not exceed -0.1% and -0.8%, we can set up the following inequalities:

-0.4 > -0.1% = -0.001
-0.4 > -0.8% = -0.008

Solving the first inequality:

-0.4 > -0.001
0.4 < 0.001

This inequality is not possible, so we can ignore it.

Solving the second inequality:

-0.4 > -0.008
0.4 < 0.008

This inequality is possible, so we can use it to determine the minimum sample mass.

0.4 < 0.008M

Dividing both sides by 0.008:

0.4 / 0.008 < M
50 < M

Therefore, the minimum sample mass that should be taken is 50 grams.

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