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3 years ago

Classify each sequence as arithmetic, geometric or neither

Mathematics

1 answer:

Karolina [17]3 years ago

3 0

Answer:

Geometric sequence = ⁵/9, ⁵/3, 5, 15, 45. . .

Arithmetic sequence = 16, 11, 6, 1, -4 . . .

Step-by-step explanation:

A geometric sequence is a sequence of numbers, in which every number in the sequence after the first term is multiplied by a common ratio, which gives us the next term. The common ratio is the ratio of a term to the term before it.

On the other hand, an arithmetic sequence is a sequence of number have a common difference between each consecutive terms. This common difference is gotten by subtracting a term from the term before it.

Now let's examine each sequence given:

==>16, 11, 6, 1, -4 . . .

11 - 16 = -5

6 - 11 = -5

1 - 6 = -5

-4 - 1 = -5

This is an arithmetic sequence. The commons difference "-5" is the same

==>⁵/9, ⁵/3, 5, 15, 45. . .

⁵/3 ÷ ⁵/9 = ⁵/3 × ⁹/5 = ⁹/3 = 3

5 ÷ ⁵/3 = ⁵ × ⅗ = 3

15 ÷ 5 = 3

45 ÷ 15 = 3

The common ratio "3" is the same. This is a geometric sequence.

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

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Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

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And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

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$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

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3 years ago

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Step-by-step explanation:

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3 years ago

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dexar [7]

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