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VikaD [51]
2 years ago
15

Help please i have a map test :)

Mathematics
1 answer:
just olya [345]2 years ago
4 0
I think it could be E
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all sides of a triangle are equal

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(b)<br> 4x+3<br> 2x-1<br> - 2 =<br> 6x+2<br> 2x-1
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If a coin is flipped 10 times, what is the probability that it will show all heads or all tails? A 1/2 B 1/512 C 1/1022 D 1/1024
Ksju [112]

The probability of multiple events happening is calculated by multiplying the probabilities of each event together. This is the case if each event is independent of the other events.

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4 0
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The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%
vagabundo [1.1K]

Answer:

1) \chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313

2) p_v =P(\chi^2_{3}>16.313)=0.000978

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}

Where O rpresent the observed values and E the expected values.  

State the null and alternative hypothesis

Null hypothesis: The distribution  is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution  is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

E = \% Grand total

E_{Catfish}=500*0.3=150

E_{Bass}=500*0.15=75

E_{Bluegill}=500*0.4=200

E_{Pike}=500*0.15=75

Part 1: Calculate the statistic

\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313

\chi^2 =16.313

Calculate the critical value

First we need to calculate the degrees of freedom given by:

df= (categories-1)=(4-1)= 3

Since the confidence provided is 95% the significance would be \alpha=1-0.95=0.05 and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be \chi^2_{crit}=7.815

We can calculate also the p value:

p_v =P(\chi^2_{3}>16.313)=0.000978

And we got the same decision reject the null hypothesis at 5% of significance.

5 0
2 years ago
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