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Rufina [12.5K]
3 years ago
5

Solve the following expression when a = 4 and c = 3 4a + 12 – 2c + c Order of operation

Mathematics
2 answers:
lbvjy [14]3 years ago
8 0

Answer:

19

Step-by-step explanation:

4×4= 16+12= 28

2×3= 6+3= 9

28-9= 19

lianna [129]3 years ago
8 0
4x4 = 16 16 + 12 = 28 2 x 3 = 6 6 + 3 = 9

28 - 9 = 19
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A researcher wishes to see if the five ways (drinking decaffeinated beverages, taking a nap, going for a walk, eating a sugary s
e-lub [12.9K]

Answer:

\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833

p_v = P(\chi^2_{4} >13.833)=0.00785

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the  drowsiness are NOT equally distributed among office workers .

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".

Assume the following dataset:

Method   Beverage   Nap   Walk   Snack   Other

Number       21             16       10         8         5

The total on this case is 60

We need to conduct a chi square test in order to check the following hypothesis:

H0: Drowsiness are equally distributed among office workers

H1: Drowsiness IS NOT equally distributed among office workers

The level of significance assumed for this case is \alpha=0.1

The statistic to check the hypothesis is given by:

\chi^2 = \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{total}{5}

And the calculations are given by:

E_{Beverage} =\frac{60}{5}=12

E_{Nap} =\frac{60}{5}=12

E_{Walk} =\frac{60}{5}=12

E_{Snack} =\frac{60}{5}=12

E_{Other} =\frac{60}{5}=12

And the expected values are given by:

Method   Beverage   Nap   Walk   Snack   Other

Number       12             12       12         12         12

And now we can calculate the statistic:

\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833

Now we can calculate the degrees of freedom for the statistic given by:

df=(catgories-1)=(5-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4} >13.833)=0.00785

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the  drowsiness are NOT equally distributed among office workers .

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Answer:

-1

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