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lorasvet [3.4K]
3 years ago
6

If an=5n-3 find the common difference​

Mathematics
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

d = 5

Step-by-step explanation:

The common difference d of an arithmetic sequence is

d = a₂ - a₁ = a₃ - a₂ = a_{n} - a_{n-1}

Given

a_{n} = 5n - 3

Generate the first 2 terms of the sequence by substituting n = 1, n = 2

a₁ = 5(1) - 3 = 5 - 3 = 2

a₂ = 5(2) - 3 = 10 - 3 = 7

d = 7 - 2 = 5

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Find a fraction in between 21/34 and 34/55
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A fraction between that is 2311/3740

Step-by-step explanation:

This is because the denominator is simplified into one number and the fraction is taken from there.

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
nlexa [21]

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

a. 11.070

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:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

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

4 0
3 years ago
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Answer:

0

Step-by-step explanation:

We have 2 points on a line so we can use the slope formula

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   = (4-4)/(-10-7)

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It would be: 12/75 = 4/25

So, your answer is 4/25

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