Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (anonymous):

A courier company claims in its advertisement that it can get all packages from city A to city B in exactly 12 hours; no more, no less. Which of the following does NOT represent the company's claim? A. The range of delivery times is zero B. The standard deviation of delivery times is zero. C. The mean and median of delivery times are both 12 hours. D. The mode of delivery times is 12 hours.

OpenStudy (anonymous):

@whpalmer4

OpenStudy (anonymous):

All deliveries are exactly 12 hours, so there is no range. The longest time is 12 hours and the shortest time is 12 hours. There is no variation. So, range is zero and standard deviation is zero, so A and B are true. The mean time would be 12 hours, since all deliveries are 12 hours. The median is also 12 hours, because if you arrange a bunch of 12s together and cross them off in pairs, you'll end up with a 12. So C is true. The mode is the most common delivery time. Since all deliveries are 12 hours, the most common time is 12 hours. So D is true. By my inspection they're all true.

OpenStudy (anonymous):

that was my problem, was im unsure of what exactly the answer is. ugh. this is just a filler class and i swear this is one of the hardest ive taken.

OpenStudy (anonymous):

Maybe someone else will have a different view. To me it seems like a bad question.

OpenStudy (anonymous):

kinda why i posted it, hoping to see what others thought. dont blame you for thinking the same thing as me.

OpenStudy (anonymous):

@jim_thompson5910 any ideas?

OpenStudy (whpalmer4):

I would say that the last choice might be the answer. You could have a mode of 12 without all deliveries taking exactly 12 hours, right?

OpenStudy (whpalmer4):

but I'll agree that it seems like a sketchy question...

OpenStudy (anonymous):

Yes, but we know that all deliveries are exactly 12 hours, so we know the mode is 12.

OpenStudy (whpalmer4):

right, I'm just saying that one could have a set of delivery times with a mode of 12 without implying that they were all 12

OpenStudy (anonymous):

Yea, thats definitely true, but does that make it the answer to the question? I don't really think so

OpenStudy (whpalmer4):

If I say that my performance on some task has a mode of 12, that's different than saying that it is always 12

OpenStudy (anonymous):

im just so touchy with my homework assignments, i hate when they do this cause i dont want to get it wrong because theyre mean lol. but i don't think the last one would be right, i was personally thinking that maybe C would be the correct answer, but i just ... dont know.

OpenStudy (anonymous):

@whpalmer4 You could say that about any of the answer choices. You could have a mean or median of 12 without them all being 12. You could have a range of 0 without them all being 12 (they could all be 11, or 10, etc) and you could have a standard deviation of 0 without them all being 12. So I say the question is bogus.

OpenStudy (whpalmer4):

well, I don't see an answer choice e) this question is bogus :-)

OpenStudy (anonymous):

^ i like the way you think. im just not going to answer it out of rebellion. then ill yell at them on monday.

OpenStudy (anonymous):

Fight the oppression of bogus math questions, on behalf of all of us

OpenStudy (anonymous):

seriously though, thank you guys for atleast trying to help. i really appreciate that. ill have some more with this ridiculous class im taking. lol. harder than i thought and its a tenth grade class and im in half college courses! mostly english though (; thanks guys!!!

OpenStudy (whpalmer4):

report back with the "correct" answer, okay?

OpenStudy (anonymous):

will do!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!