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andriy [413]
3 years ago
10

Let A ⊆ B ⊆ C be rings. Suppose C is a finitely generated A-module. Does it follow that B is a finitely generated A-module?

Mathematics
1 answer:
Lunna [17]3 years ago
8 0

Answer:

Let A ⊆ B ⊆ C be rings. If C is a finitely generated A-module. Then B is a finitely generated A-module.

Step-by-step explanation:

Draw a ring and call it A, then draw another circle with a longer radius from the same centre of A and call it B, then draw another from the same centres of A and B, but with the longest radius and call it C.

Then, when you say A ⊆ B ⊆ C, this means that A is a subset of and equal to B which is a is a subset of and equal to C.

Meaning:

1. A is in B and B is in C.

2. The values in A are the only values in B. i.e

If A = {2,4,6} then B = {2,4,6}

3. The values of ring B are the only values in ring C. i.e. if B = {2,4,6} then C = {2,4,6}.

4. There is no more values in B that is not in A.

5. There are no more values in C that is not in B.

Since they are subsets of each other defined by ⊆, which makes the subset exactly the same as the host set or superset.

So the same rule that applies to C will apply to B

A finitely generated module is a module that has a finite generating set. A finitely generated module over a ring A may also be called a finite A-module.

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Answer:

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Step-by-step explanation:

Volume of Sphere = \frac{4}{3}\pi r^3

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V = (4)(3.14)(11,560,000)/3

V = 48,422,415 km³

<u><em>Upto 3 sfs:</em></u>

V = 4.84 × 10⁷ km³

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The area of a square garden is 64 m2. How long is each side of the<br> garden?
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Step-by-step explanation:

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3 years ago
The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a sta
nordsb [41]

Answer:

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce?

This is the pvalue of Z when X = 8.2 + 0.02 = 8.22 subtracted by the pvalue of Z when X = 8.2 - 0.02 = 8.18. So

X = 8.22

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.22 - 8.2}{0.018}

Z = 1.11

Z = 1.11 has a pvalue of 0.8665

X = 8.18

Z = \frac{X - \mu}{s}

Z = \frac{8.18 - 8.2}{0.018}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

8 0
4 years ago
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