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kow [346]
3 years ago
13

Plsss help meee 45 points plus brainliest

Mathematics
1 answer:
romanna [79]3 years ago
5 0

<em>Hello There!!</em>

Let us call X the equal distances GH = HI = IJ

GH = X

HI = X

IJ = X

Note that HJ = HI + IJ, so

HJ = X + X

10 = 2X

X = 5

Well, We're looking at figure we can conclude that:

GJ + JK = GK

GJ = GH + HI + IJ

GJ = 15

GJ + JK = GK

15 + JK = 24

JK = 9

GH = X = 5

HI = X = 5

IJ = X = 5

JK = 9

HopeThisHelpsYou!

_{Losersbrazts}

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Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2  CI 95 %  =  (  0,056 ;  0,164 )

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n₁  =  500

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Sample information for people over 18

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Hypothesis Test

Null hypothesis                        H₀              p₁  =  p₂

Alternative Hypothesis           Hₐ              p₁  ≠  p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

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To calculate   z(s)

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EED = √(p₁*q₁)n₁  +  (p₂*q₂)/n₂

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EED = √0,00046  +  0,0003125

EED = 0,028

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Then

z(s)  =  0,11 / 0,028

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Q2  CI  95 %   =  (  p₁  -  p₂  ) ±  z(c) * EED

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CI 95%  = (  0,11  ±  0,054 )

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