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vodomira [7]
3 years ago
13

Which dot plot represents this data set?

Mathematics
2 answers:
Bumek [7]3 years ago
3 0

Answer:

Dot plot D

Step-by-step explanation:

Dot plot D represents the dot plot with one dot per each number in the data set.

kodGreya [7K]3 years ago
3 0
Dot D
Because theirs no 13
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Use distance formula d=√(X2 – X1 )2 + (X2 – X1 )2, calculate the distance r from the origin to the point r (-2, √5).
lesya692 [45]
d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \\d = \sqrt{(\sqrt{5} - (-2))^{2} + (\sqrt{5} - (-2))^{2}} \\d = \sqrt{(\sqrt{5} + 2)^{2} + (\sqrt{5} + 2)^{2}} \\d = \sqrt{(4 + 4\sqrt{5} + 5) + (4v + 4\sqrt{5} + 5)} \\d = \sqrt{0 + 0} \\d = \sqrt{0} \\d = 0
7 0
3 years ago
PLEASE HELPPP!!! I will mark brainlist​
Maslowich

Answer:

d

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Divide. Yes rectangular model to record the partial quotient <br> 246/3=
ra1l [238]
82... because 24/3 is 8 and 6/3 is 2
Therefore you get 82
5 0
3 years ago
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
Help ..........................................​
nalin [4]

Answer:

c. 45d+.25m

Step-by-step explanation:

3 0
3 years ago
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