anonymous
  • anonymous
Ethan rolls a 6-sided die. What is the probability that he gets a number greater than 4 or an even number? A. B. C. D.
Geometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1/3 2/3 1/6 5/6
OpenStudy9
  • OpenStudy9
B
OpenStudy9
  • OpenStudy9
Let me tell you why

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More answers

anonymous
  • anonymous
ok
anonymous
  • anonymous
thats my name btw :0
mathstudent55
  • mathstudent55
See how many of the 6 possible outcomes are the ones you want.
anonymous
  • anonymous
thanks everyone
FortyTheRapper
  • FortyTheRapper
How many numbers on the dice are greater than 4? 3 How many of them are even AND less than 4? 1 Add those two answers and put it over 6. Simplify and the answer is there
OpenStudy9
  • OpenStudy9
it is 2,4,5, and 6 because greater than 4(5 and 6) and (even(2,4,6) 6 repeats, so 2,4,5,6 is 4/6=2/3
anonymous
  • anonymous
i got 2/3
anonymous
  • anonymous
thank you
OpenStudy9
  • OpenStudy9
welcome!
anonymous
  • anonymous
i have another one just like it can u help me with it?
mathstudent55
  • mathstudent55
|dw:1449528014855:dw|
mathstudent55
  • mathstudent55
2/3 is correct.
anonymous
  • anonymous
You roll a 6-sided die with faces numbered 1 through 6, and toss a coin. What is the probability of rolling a 5 or getting tails? A. B. C. D.
anonymous
  • anonymous
1/3 7/12 3/4 1/12
anonymous
  • anonymous
@mathstudent55 @FortyTheRapper
anonymous
  • anonymous
1/12?
FortyTheRapper
  • FortyTheRapper
So, you only want to roll a 5, which is 1 number. So put 1 over 6 to get 1/6. That's the dice possibility. Ofc, two sides to a coin so it's always 50/50, or 1/2. Then because you're doing both, you would multiply the denominators to get 12 I would first put them both in terms of 12, so it will be easier 1/12 and 6/12, instead of 1/6 and 1/2 So, now add the 1/12 and 6/12 together to get your answer
anonymous
  • anonymous
7/12
FortyTheRapper
  • FortyTheRapper
Yep ^
anonymous
  • anonymous
thanks
OpenStudy9
  • OpenStudy9
wait 2kende are you here still?
OpenStudy9
  • OpenStudy9
I think I caught a mistake
anonymous
  • anonymous
yes
anonymous
  • anonymous
wut mistake
OpenStudy9
  • OpenStudy9
1/6 is 2/12 and 1/2 is 6/12
anonymous
  • anonymous
damn.. ):but its too late i already submitted the question
OpenStudy9
  • OpenStudy9
ok
anonymous
  • anonymous
can you help me with this next one?
OpenStudy9
  • OpenStudy9
sure
anonymous
  • anonymous
Find the missing probability in the table below. X 1 2 3 4 P(X) 0.40 0.20 0.15 ? A. 0.25 B. 0.75 C. 0.85 D. 0.15
FortyTheRapper
  • FortyTheRapper
@OpenStudy9 Ahh, I just noticed it also. Good find. My mistake @2kende
OpenStudy9
  • OpenStudy9
So there is a pattern in this question
anonymous
  • anonymous
yes
OpenStudy9
  • OpenStudy9
first, subtract .2, then .5
anonymous
  • anonymous
i added them all together and got .75
OpenStudy9
  • OpenStudy9
sorry, .05
OpenStudy9
  • OpenStudy9
the question asks for the missing value right?
anonymous
  • anonymous
yes, the missing probability
OpenStudy9
  • OpenStudy9
IS that all in the question?
anonymous
  • anonymous
yes
OpenStudy9
  • OpenStudy9
It could be a quadratic, lt me take a closer look
OpenStudy9
  • OpenStudy9
are those the right numbers?
OpenStudy9
  • OpenStudy9
Is this "quiz" timed?
anonymous
  • anonymous
no
anonymous
  • anonymous
i think it could be D
OpenStudy9
  • OpenStudy9
Can you come back to problems? If not, go with D
anonymous
  • anonymous
no i cant go back
OpenStudy9
  • OpenStudy9
the go with D for now;)
anonymous
  • anonymous
ok, do u have time for a couple more?
OpenStudy9
  • OpenStudy9
maybe one more
OpenStudy9
  • OpenStudy9
What is it?
anonymous
  • anonymous
Given the table of candy distributions below, find P(Not orange), the probability of selecting anything but an orange candy at random from a large bag. Red Orange Yellow Green Blue Brown 0.13 0.20 ? 0.16 0.24 0.13 A. 0.80 B. 0.20 C. 0.70 D. 0.50
OpenStudy9
  • OpenStudy9
You typed this very confusing
OpenStudy9
  • OpenStudy9
Where is the question mark?
anonymous
  • anonymous
the question mark is the missing yellow color
anonymous
  • anonymous
thats what were trying to find
OpenStudy9
  • OpenStudy9
ok, the rest go as folows, got it, one sec..
OpenStudy9
  • OpenStudy9
A
OpenStudy9
  • OpenStudy9
Let me tell you why
OpenStudy9
  • OpenStudy9
All must add up to 1
OpenStudy9
  • OpenStudy9
yellow is .14
OpenStudy9
  • OpenStudy9
orange is .2
OpenStudy9
  • OpenStudy9
1-.2 is .8
OpenStudy9
  • OpenStudy9
OK?
OpenStudy9
  • OpenStudy9
You still there?
anonymous
  • anonymous
ohhh i get it
anonymous
  • anonymous
thanks mann(:
OpenStudy9
  • OpenStudy9
Your welcome! No biggie, but do you mind giving me a medal?
anonymous
  • anonymous
1 more please?
anonymous
  • anonymous
and sure
OpenStudy9
  • OpenStudy9
Sure!
OpenStudy9
  • OpenStudy9
Bring it on!
anonymous
  • anonymous
lol ok
anonymous
  • anonymous
Given the Venn diagram below, if a student is randomly selected, what is the probability that he or she is not attending both colleges? A. 86% B. 82% C. 80% D. 84%
anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/54c0b02be4b0e0dd17c94541-kianatalani-1421914222284-venndiagram.gif
OpenStudy9
  • OpenStudy9
I think i almost got it..
OpenStudy9
  • OpenStudy9
A again
OpenStudy9
  • OpenStudy9
Here's why
OpenStudy9
  • OpenStudy9
28 are attending both colleges
OpenStudy9
  • OpenStudy9
They all add up to 200
OpenStudy9
  • OpenStudy9
(200-28)/200=.86 Hence, 86 percent
OpenStudy9
  • OpenStudy9
OK?
anonymous
  • anonymous
ohh okk
anonymous
  • anonymous
i got a 90% <3 thanks
OpenStudy9
  • OpenStudy9
Your welcome!
OpenStudy9
  • OpenStudy9
probably, the one that the forty rapper guy missed, and 2 others
anonymous
  • anonymous
lol yea, but im glad u came through (:

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