anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Directrix
  • Directrix
#1) Agree - Fundamental Principle of Counting #2) Disagree on CD Problem #3) Disagree on mean problem #4) Agree on mode problem (Bimodal)
Directrix
  • Directrix
#5) Disagree on Median Problem The median is the middle data point after the data is arranged in increasing order of size. 9 is not in the middle. It is the largest data point.
crabbyoldgamer
  • crabbyoldgamer
I disagree with #1. There isn't enough info given. What if some of the shirt's colors clash with pants colors? Dressing isn't a random event.

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More answers

anonymous
  • anonymous
would the answer to number 5 be A?
Directrix
  • Directrix
#1 is correct. The number of ways a person can make an outfit is without regard to whether the outfits match.
Directrix
  • Directrix
The median is not 3.4. Do this: Post the 7 numbers in order from smallest to largest. @kittymeow101 Then, we will find the one in the middle which is the median.
crabbyoldgamer
  • crabbyoldgamer
"#1 is correct. The number of ways a person can make an outfit is without regard to whether the outfits match." Not in the real world, kiddo.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
1.2,3.4, 3.4,5.6, 5.6, 7.8,9.0
anonymous
  • anonymous
the answer would be 5.6
Directrix
  • Directrix
>the answer would be 5.6 Correct
anonymous
  • anonymous
would the answer to the CD problem be 70?
Directrix
  • Directrix
On the mean problem, you can add them and divide by 11 or enter them here and let the program calculate the mean. http://www.alcula.com/calculators/statistics/mean/
anonymous
  • anonymous
would the answer to the mean problem be 20.63?
Directrix
  • Directrix
>would the answer to the CD problem be 70? That is what I got. Order does not matter here. The number of ways to pick a group of 4 from 8 is C(8,4) = 70.
Directrix
  • Directrix
>would the answer to the mean problem be 20.63? That is what I got. Yes.
Directrix
  • Directrix
That is it for this thread, right?
anonymous
  • anonymous
yes ty

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