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SIZIF [17.4K]
2 years ago
5

forty volunteer drivers are separated into two groups of 20 drivers each at random. The first group is asked to pay particular a

ttention to braking smoothly when approaching a red light or stop sign. The second group is given no special instructions. All drivers report, at the end of the month, the gallons of gasoline used. The mean for the first group is 43.16 gallons of gas. The mean for the second group is 54.63 gallons of gas. The difference in means for the two groups is -11.47 gallons of gas. The data for both groups are combined and redistributed at random into two groups of 20. The means for each redistributed group are calculated, and the difference in means is recorded. The differences are represented in the histogram. Which of these conclusions and justifications is the most reasonable?
Mathematics
1 answer:
ale4655 [162]2 years ago
4 0

The reasonable conclusion is that the randomization distribution provides strong evidence that the difference in the means for the two treatments group is due to the treatment because the difference is highly unlikely based on random assignment error.

<h3>How to explain the information?</h3>

Given the mean for the first group is 43.16 gallons of gas and that the mean for the second group 54.63 gallons of gas. The difference in the means for the two groups is - 11.47 gallons of gas.

Here, the data for both groups are combined and redistributed at random into two groups of 20. The histogram shows that the distribution is approximately symmetric. Therefore, the randomization technique can be used.

Learn more about histogram on:

brainly.com/question/2962546

#SPJ1

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Answer:

\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

Step-by-step explanation:

We want to find \lim_{x \to 2^-} \frac{|x-2|}{x-2}.

By definition:

|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1).

The limit of a constant is the constant.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

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