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Ede4ka [16]
3 years ago
6

The map of a neighborhood park is shown in the diagram. Myra is walking from the pool to the fishing pond along the diagonal sho

wn. To the nearest yard, how far will Myra walk?

Mathematics
1 answer:
frozen [14]3 years ago
4 0
If you have ever done any walking, you know that the diagonal distance is more than the length of either side and less than their sum. The only selection in that range is ...
   C)   119 yards

_____
If you want to actually figure it out, you would use the Pythagorean theorem.
   distance = √(100² + 65²) = √14225 ≈ 119.269 . . . . yards
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What is the answer to 43.26 ÷ 14​
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Step-by-step explanation:

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in geotown, there are 210 own a tablet. This is at least 4/5 of all teenagers that live in geotown. What is the maximum number o
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210 = 4/5 (This means that 4 out of five people (210 people) are a teenager)

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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 .

7 0
4 years ago
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