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andrey2020 [161]
3 years ago
15

jessica is hiking in the mountains with her llama, her dog and her cat she comes to river where she must help each animal across

one at a time she has a problem she can't leave llama along with the dog, and she can't leave a cat alone with the dog, what is the fewest trips Jessica can take to get the llama the dog and the cat the other side of the river
Mathematics
2 answers:
Wittaler [7]3 years ago
8 0
She takes the dog first, then the llama, then the cat. Easy. 3 trips hope I helped
Alexxandr [17]3 years ago
8 0
First take the dog to the other side. Then take the llama across. When you get to where the dog is, drop the llama off and pickup the dog again. Take the dog to where the cat is. Drop off the dog, pick up the cat. Take the cat to the other side of the river, leave it with the llama. Go back, pick up the dog, and continue on! You need 4 round trips, or 8 one-way trips. Hope this helps you!
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<em><u>Polygon:</u></em> a figure formed by coplanar segments

with one distinct interior region

Figure A, Figure B, and Figure D would be considered polygons.

Notice that those figures I mentioned above are

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In addition, these figures don't have two separate

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Figure C would not be a polygon because

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4 0
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2. The National Safety Council routinely analyzes the benefit of seat belt use on driver safety. Their data showed that among 28
JulijaS [17]

Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let p_1 = <u><em>proportion of deaths when drivers were not wearing seat belts.</em></u>

p_2 = <u><em>proportion of deaths when drivers were wearing seat belts.</em></u>

So, Null Hypothesis, H_0 : p_1=p_2      {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, H_A : p_1\neq p_2     {means that there is difference in the proportion of deaths between the 2 groups}

The test statistics that would be used here <u>Two-sample z test for proportions;</u>

                          T.S. =  \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 deaths when drivers were not wearing seat belts = \frac{31}{2823} = 0.011

\hat p_2 = sample proportion of deaths when drivers were wearing seat belts = \frac{16}{7765} = 0.002

n_1 = sample of drivers not wearing seat belts = 2823

n_2 = sample of drivers wearing seat belts = 7765

So, <u><em>the test statistics</em></u>  =  \frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }

                                       =  4.438

The value of z test statistics is 4.438.

<u>Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that there is difference in the proportion of deaths between the 2 groups.

Also, <u>Margin of error</u> (E) =  1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }

                                        =  1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }

                                        =  <u>0.00397</u>

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