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Naya [18.7K]
3 years ago
9

WILL MARK BRAINLEIST PLEASE HURRY

Mathematics
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

<h3>The reason which is incorrect is in Step 2 and reason 2</h3>

Step-by-step explanation:

Given that Logan's equation is \frac{3}{2}\times (x-10)=50

<h3>To simplify and find the which reason is incorrect :</h3>

Logan's solution and reasoning for solving an equation are shown below:

Step1 \frac{3}{2}\times (x-10)=50

Reason 1: Given

Step2 \frac{3}{2}\times x-10+10=50+10

Reason 2: Addition Property of Equality

Step3 \frac{3}{2}\times x=60

Reason 3: Simplify

Step4 \frac{3}{2}\times x \times (\frac{2}{3})=60\times (\frac{2}{3})

Reason 4: Division Property of Equality

Step5 x=40

Reason 5: Simplify

<h3>The reason which is incorrect is in Step 2 and reason 2</h3><h3>The corrected steps are</h3>

Step1 \frac{3}{2}\times (x-10)=50

Reason 1: Given

Step2 \frac{3}{2}\times (x)-\frac{3}{2}\times (10)=50

Reason 2: Distributive property

Step3 \frac{3}{2}\times x-15=60

Reason 3: Simplify

Step4 \frac{3}{2}\times x-15+15=50+15

Reason 4: Addition Property of Equality

Step5 \frac{3}{2}\times x=65

Reason 5: Simplify

Step6 \frac{3}{2}\times x \times (\frac{2}{3})=65\times (\frac{2}{3})

Reason 6: Division Property of Equality

Step7 x=\frac{130}{3}

Reason 7: Simplify

<h3 />
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6 0
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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
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Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
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