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Digiron [165]
3 years ago
13

A silversmith combined pure silver that cost $30.48 per ounce with 52 oz of a silver alloy that cost $21.35 per ounce. How many

ounces of pure silver were used to make an alloy of silver costing $24.76 per ounce?
oz
Mathematics
1 answer:
weqwewe [10]3 years ago
6 0

Let x be the amount (in oz) of pure silver used in this alloy. Each ounce of the pure silver contributes a value of $30.48, so that the total cost of the amount of pure silver that gets used is $30.48x.

The same goes for the secondary alloy, where each ounce costs $21.35, so that 52 ounces of the alloy contributes a total cost of $21.35*52.

The final alloy has a total mass of (x + 52) ounces, and each ounce costs $24.76, so that this alloy's totat cost is $24.76*(x + 52).

So we have

30.48x + 21.35*52 = 24.76*(x + 52)

30.48x + 1110.20 = 24.76x + 1287.52

5.72x = 177.32

=> x = 31

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Diano4ka-milaya [45]

Considering the Central Limit Theorem, we have that:

a) The probability cannot be calculated, as the underlying distribution is not normal and the sample size is less than 30.

b) The probability can be calculated, as the sample size is greater than 30.

<h3>What does the Central Limit Theorem state?</h3>

It states that the sampling distribution of sample means of size n is approximately normal has standard deviation s = \frac{\sigma}{\sqrt{n}}, as long as the underlying distribution is normal or the sample size is greater than 30.

In this problem, the underlying distribution is skewed right, that is, not normal, hence:

  • For item a, the probability cannot be calculated, as the underlying distribution is not normal and the sample size is less than 30.
  • For item b, the probability can be calculated, as the sample size is greater than 30.

More can be learned about the Central Limit Theorem at brainly.com/question/16695444

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