Continued from the previous question. The number of trips each week that take between 70 and 82 minutes is:

Question

BlackTom AI · Accepted Answer

The prompt for this question is incomplete in a crucial way: the answer options are missing. Without the set of possible choices, we cannot map any computed result to a labeled option or verify which lettered choice corresponds to which numeric value.

What we do know is the task: determine the number of trips per week that have durations between 70 and 82 minutes. To perform this calculation in a typical scenario, you would need one of the following sources of information:
- A data table or dataset listing each trip’s duration, from which you would count how many trips fall in the inclusive interval [70, 82].
- A summary distribution (for example, a histogram, frequency table, or a probability distribution) that provides counts or probabilities for each duration bin, from which you could sum the counts for bins that intersect with 70–82 minutes.
- If a model or approximation is being used (for instance, assuming trips durations follow a normal distribution with known mean and standard deviation), you would compute the expected number of trips in that interval as: total_trips × P(70 ≤ duration ≤ 82) using the appropriate parameters μ and σ and the standard normal CDF.

Given the absence of answer options, here are practical steps you could take to compute the value once you have the data:
- If you have raw data: filter the dataset to include only trips with duration between 70 and 82 minutes, then count the number of such trips. If you’re asked for a weekly count, ensure the data is segmented by week and sum the qualifying trips per week.
- If you have a frequency table: identify all bins that lie between 70 and 82 minutes, sum their frequencies, and, if needed, adjust for any partial bins at the interval boundaries depending on how the data is binned.
- If you’re working with a normal approximation: verify that the assumption is appropriate, compute z-scores for 70 and 82, look up the standard normal probabilities between those z-scores, multiply by the total weekly trips, and round as instructed by your course or dataset conventions.

Because the provided options are empty, there is no way to decide which option corresponds to 163 or any other value. To proceed accurately, you would need the complete list of answer choices or access to the underlying data or distribution used to generate the weekly counts.

If you can share the missing answer options, or provide the dataset or distribution parameters (mean, standard deviation, and total weekly trips), I can walk through the exact calculation and map the result to the correct option with a clear justification for each choice.

Continued from the previous question. The number of trips each week that take between 70 and 82 minutes is:

View Explanation

Log in for full answers

Similar Questions

位置24的问题 In constructing a frequency distribution, as the number of classes is decreased, the class width can increase or decrease depending on the data values.TrueFalse

Assuming you have data for a variable with 2,000 values, using the 2k ≥ n guideline, what is the least number of groups that should be used in developing a grouped data frequency distribution?

The following class limits would be acceptable for developing a frequency distribution on income: $0 - $5,000 $5,001 - $10,000 $10,001 - $20,000 Over $20,000

A common rule of thumb for determining how many classes to use when developing a frequency distribution with classes is:

Continued from the previous question. The number of trips each week that take between 70 and 82 minutes is:

Continued from the previous question. The number of trips each week that take between 70 and 82 minutes is:

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler