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3 years ago

Que son las figuras planas

Mathematics

1 answer:

RideAnS [48]3 years ago

5 0

Figuras planas son dos dimensiones formas como cuadrados, triangulos y circulos

↑ ↑ ↑ Espero que esto ayude! :D

In English, this means --

Flat figures are two-dimensional shapes like squares, triangles and circles

↑ ↑ ↑ Hope this helps! :D

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4 years ago

Classify each pair of numbered angles.

kifflom [539]

Adjacent angles are angles with a common side and vertex, linear pairs are adjacent angles that are supplementary, and vertical angles are angles made by the same two lines but on opposite sides.

For the first one, 5 and 6 clearly do not share a side but they are made up by the same 2 lines and are opposite of each other, making them vertical.

For the next one, since the angles only share 1 line (and not a side) they can't be any of the above.

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3 years ago

Read 2 more answers

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

2 3/4 + 4 2/3 = <br> 8 1/3 + 8 5/7 =

satela [25.4K]

7 5/12 or 7.41
17 1/21 or 17.04

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3 years ago

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Wewaii [24]

Answer:

6 * (m + n)

Step-by-step explanation:

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3 years ago

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