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enot [183]
3 years ago
9

The base of a cube is parallel to the horizon. If the cube is cut by a plane to form a cross section, under what circumstance ca

n the cross section be a non-rectangular parallelogram?
when the plane cuts three faces of the cube, separating one corner from the others

when the plane passes through a pair of vertices that do not share a common face


when the plane is perpendicular to the base and intersects two adjacent vertical faces


when the plane makes an acute angle to the base and intersects three vertical faces


not enough information to answer the question
Mathematics
2 answers:
Nikolay [14]3 years ago
7 0
Based on the given question above, the correct answer would be the fourth option. The <span>circumstance that the cross section can be a non-rectangular parallelogram is that, </span><span>when the plane makes an acute angle to the base and intersects three vertical faces. Hope this answers the question. </span>
Vaselesa [24]3 years ago
3 0

Answer:

the fourth option

Step-by-step explanation:

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Answer:

z=\frac{0.616-0.5}{\sqrt{\frac{0.5(1-0.5)}{297}}}=3.998  

p_v =2*P(z>3.998)=0.0000639  

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Step-by-step explanation:

Information given  

n=297 represent the random sample of male taken

X=183 represent the  men who said yes, they had driven a car when they probably had too much alcohol

\hat p=\frac{183}{297}=0.616 estimated proportion of men who said yes, they had driven a car when they probably had too much alcohol

p_o=0.5 is the value that we want to test

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Hypothesis to test

We need to conduct a hypothesis in order to test the claim that the majority of men in the population (that is, more than half) would say that they had driven a car when they probably had too much alcohol, and the system of hypothesis are:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

The statistic is given by:

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

After replace we got:

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

z=\frac{0.616-0.5}{\sqrt{\frac{0.5(1-0.5)}{297}}}=3.998  

Decision

We have a right tailed test so then the p value would be:  

p_v =2*P(z>3.998)=0.0000639  

With the most common significance levels used \alpha= 0.1, 0.05, 0.01 we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5

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