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bazaltina [42]
4 years ago
8

Could anyone please help me with this question? thanks

Mathematics
1 answer:
Tpy6a [65]4 years ago
3 0
AB = AC, so this is an isosceles triangle, then angle B=angle C =66⁰.
m∠BAD = m∠CAD, so AD is a bisector. 
In an isosceles triangle, bisector is also median and height.
So AD is perpendicular bisector.
So angle x = 90⁰. 

In triangle ABC
m∠A+m∠B+m∠C=180⁰
m∠A+66+66=180
m∠A=180-66-66=48⁰

m∠A=48⁰
AD is a bisector of A, so y=m∠A/2=48⁰/2=24⁰


x=90⁰, y=24⁰
Answer is A
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Step-by-step explanation:

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A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selecti
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Answer:

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    Alternate Hypothesis, H_A : p > 0.50

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed.

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Step-by-step explanation:

We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is greater than 0.5.

Assume that sample data consists of 78 girls in 144 ​births.

Let p = <u><em>population proportion of baby girls</em></u>

(a) So, Null Hypothesis, H_0 : p = 0.50     {means that the proportion of baby girls is equal to 0.50}

Alternate Hypothesis, H_A : p > 0.50     {means that the proportion of baby girls is greater than 0.50}

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.

(e) The test statistics that would be used here <u>One-sample z test for proportions</u>;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of baby girls =  \frac{78}{144} = 0.54

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So, <u><em>the test statistics</em></u>  =  \frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }  

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The value of z test statistics is 0.96.

(f) <u>The P-value of the test statistics is given by;</u>

            P-value = P(Z > 0.96) = 1 - P(Z < 0.96)

                          = 1 - 0.8315 = 0.1685

<u></u>

(g) <u>Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.</u>

(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the proportion of baby girls is equal to 0.50.

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Step-by-step explanation:

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