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Linear Algebra 15 Online
OpenStudy (anonymous):

CHECK PLEASE! Here are the ages of the 10 finalists in a baking competition. 61 52 52 74 83 30 38 49 58 60 Which modified box-and-whisker plot correctly displays the data? C?

OpenStudy (anonymous):

i think its C, but im not sure.

HanAkoSolo (jamierox4ev3r):

@CatalystDuzIt try not to provide answers right off the bat...thanks! We're here to teach, not to give free answers (that's google's job!) and @CrazyCountryGirl , to eliminate answers, find the lowest outlier and the highest outlier

HanAkoSolo (jamierox4ev3r):

give me a sec while i look at the choices

OpenStudy (anonymous):

ok lol

HanAkoSolo (jamierox4ev3r):

then, you take all the values and find the average by mean

OpenStudy (anonymous):

@jamierox4ev3r...Does that mean I'm wrong? I kind of just wanted someone to check it first...

HanAkoSolo (jamierox4ev3r):

OH SHOOT I didn't realize you put your guess there xD oops lol but yes, there is a bit of error in your thinking

HanAkoSolo (jamierox4ev3r):

to make these graphs, you have to find the median of the set. but first you must order everything, like this:|dw:1424572978523:dw|

HanAkoSolo (jamierox4ev3r):

so what's the median? @CrazyCountryGirl

OpenStudy (anonymous):

@jamierox4ev3r Median = 55

HanAkoSolo (jamierox4ev3r):

good. then find the medians of the first half of the data, then the second half

HanAkoSolo (jamierox4ev3r):

let me know what numbers you get...or if you want me to explain

HanAkoSolo (jamierox4ev3r):

@CrazyCountryGirl sorry for the late-ish reply v.v

OpenStudy (anonymous):

30, 38, 49, 52, 52 - Median = 49 58, 60, 61, 74, 83 - Median = 61 @jamierox4ev3r

HanAkoSolo (jamierox4ev3r):

yep! so that's where the borders of your box would be based off from what you know now, what's the answer? @CrazyCountryGirl ?

OpenStudy (anonymous):

B?

OpenStudy (anonymous):

@Jamierox4ev3r

HanAkoSolo (jamierox4ev3r):

yep! good job :D

HanAkoSolo (jamierox4ev3r):

hope this helped :)

OpenStudy (anonymous):

It did, thanks!

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