Parking at a large university has become a very big problem. university administrators are interested in determining the average parking time (e. g. the time it takes a student to find a parking spot) of its students. an administrator inconspicuously followed 280 students and carefully recorded their parking times. which of the following graphs should not be used to display information concerning the students parking times? a. pie chartb. stem-and-leaf displayc. histogramd. box plot

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

Step-by-step explanation:

For this case our variable of interest is the average parking time of its students. And we have 280 values for these times. So then the variable of interest is quantitative.

Which of the following graphs should not be used to display information concerning the students parking times?

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

b.Stem-and-leaf display

That incorrect since the Stem and leaf plot is useful when we want to plot quantitative data.

c. Histogram

That incorrect since the Histogram is ideal when we want to plot quantitative data and analyze the distribution of the data.

d. Box plot

That incorrect since the Box plot is ideal and useful when we want to plot quantitative data and see central tendency measures.

it costs $20 to interview each person for a survey. find the cost to obtain a +- 3% margin of error.

step-by-step explanation:

copy and paste, put it in google

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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