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tangare [24]
3 years ago
5

A factory produces three kinds of juice,orange,apple and lime.Last month,6810 bottles of juice were produced.1/3 of that number

is orange juice.The number of bottles of apple juice is three times the number of bottles of lime juice.How many bottles of each type of juice were produced?
Mathematics
1 answer:
Vesna [10]3 years ago
5 0
2270 bottles of orange juice, 3027 bottles of apple juice, 1513 bottles of lime juice
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Write an equation of the line that passes through (-3,3) and is perpendicular to the line 2y=8x-6
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Answer:

y = (-1/4)x + 9/4

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2y = 8x - 6

So, y = 4x - 3  (Slope = 4)

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3 0
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The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed
algol [13]

Answer:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860

p_v = P(\chi^2_{5} >5.860)=0.32

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

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:

Number:      1,    2 ,   3 ,  4 , 5    ,6

Frequency: 27, 31, 42, 40, 28, 32

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

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The observed values are given:

O_{1}=27   O_{2}=31

O_{3}=42  O_{4}=40

O_{5}=28  O_{6}=32

The expected values are given by:

E_{1} =\frac{1}{6}*200=33.33   E_{2} =\frac{1}{6}*200=33.33

E_{3} =\frac{1}{6}*200=33.33   E_{4} =\frac{1}{6}*200=33.33

E_{5} =\frac{1}{6}*200=33.33   E_{6} =\frac{1}{6}*200=33.33

And now we can calculate the statistic:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860

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

df=Categories-1=6-1=5

And we can calculate the p value given by:

p_v = P(\chi^2_{5} >5.860)=0.32

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

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

4 0
3 years ago
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