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muminat
3 years ago
7

Which sample fairly represents the population? Select two options.

Mathematics
2 answers:
ExtremeBDS [4]3 years ago
7 0

Answer:

b) observing every person walking down Main Street at 5 p.m. one evening to determine the percentage of people who wear glasses.

d) taking a poll in the lunch room (where all students currently have to eat lunch) to determine the number of students who want to be able to leave campus during lunch.

Step-by-step explanation:

These are the two options that are most likely to give you a sample that fairly represents the population. In the first case, the sample that you obtain is likely to be a good representation because Main Street is a road where a great variety of people walk. Moreover, 5 pm is also a time that will allow you to see  a great number of different people. The second answer will also give you a good sample, as the poll would include all students in the lunch room, which is all students in the school (the whole population).

julia-pushkina [17]3 years ago
3 0

The answers are...

B) calling every third person on the soccer team’s roster to determine how many of the team members have completed their fundraising assignment

E) taking a poll in the lunch room (where all students currently have to eat lunch) to determine the number of students who want to be able to leave campus during lunch

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Answer:

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.98}{2} = 0.99. So we have T = 2.462

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The lower end of the interval is the sample mean subtracted by M. So it is 375 - 41.13 = 333.87 kWh

The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

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<h3>How to determine the remaining zeros</h3>

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