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padilas [110]
3 years ago
15

Jill converted the equation of the line 15 x minus 14 y = negative 2 into slope-intercept form and found the slope and y-interce

pt of the line as follows. 15 x minus 14 y = - 2. 15 x minus 4 y minus 15 x = - 2 minus 15 x. - 4 y = -2 minus 15 x. - 4 y Over - 4= - 2 minus 15 x Over - 4 Y = - 2 Over -4 minus 15 x Over - 4 . y = one-half minus 15 Over 4 x. y = - 15 Over 4x + one-half. slope = -15 Over 4. y-intercept = one-half. What was her mistake? She mixed up the slope and the y-intercept. She got the sign of the slope wrong. She got the sign of the y-intercept wrong. She found the reciprocals of the slope and the y-intercept.
Mathematics
2 answers:
Olegator [25]3 years ago
6 0

Answer:

B

Step-by-step explanation:

She got the sign of the slope wrong.

vova2212 [387]3 years ago
4 0

Answer:

B. She got the sign of the slope wrong.

Step-by-step explanation:

This is the correct answer for edge

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Answer:

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

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Step-by-step explanation:

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n=180 represent the random sample taken  

X=125 represent the number of americans between 17 to 24 that not qualify for the military

\hat p=\frac{125}{180}=0.694 estimated proportion of americans between 17 to 24 that not qualify for the military

p_o=0.75 is the value that we want to test  

\alpha=0.05 represent the significance level  

Confidence=95% or 0.95  

z would represent the statistic (variable of interest)  

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We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :  

Null hypothesis: p\geq 0.75  

Alternative hypothesis:p < 0.75  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.  

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

4) Statistical decision  

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The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

If we compare the p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

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