anonymous
  • anonymous
Amy has 35 tiles of the same shape and size in a box. The colors of and amounts of the tiles are shown below: 12 red tiles 8 yellow tiles 15 pink tiles Without looking in the box, Amy takes out a tile at random. She then replaces the tile and takes out another tile from the box. What is the probability that Amy takes out a yellow tile in both draws? 8 over 35 multiplied by 7 over 34 equal 56 over 1190 8 over 35 multiplied by 8 over 35 equal 64 over 1225 8 over 35 plus 7 over 34 equal 517 over 1190 8 over 35 plus 8 over 35 equal 560 over 1225
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
we have 8 favorable outcomes and 12+8+15= 35 possible outcomes so probability= favorable outcomes/ possible outcomes=...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
that is the probability associated to a single extraction
Michele_Laino
  • Michele_Laino
favorable outcomes = 8 possible outcomes = 35 \[\Large \frac{{{\text{favorable outcomes}}}}{{{\text{possible outcomes}}}} = ...\]
anonymous
  • anonymous
so i have to divide
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
the probability is 8/35 it is simple!
Michele_Laino
  • Michele_Laino
for a single extraction
anonymous
  • anonymous
so what about the other part
Michele_Laino
  • Michele_Laino
for second extraction, the probability is the same, namely it is 8/35, since Amy replaces the tile into the box. Now the requested probability is the product of the two probabilities, namely: probability = (8/35)*(8/35)=...?
anonymous
  • anonymous
64/1225
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Yay!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
A coin was flipped 150 times. The results of the experiment are shown in the following table: Heads Tails 90 60 Which of the following best describes the experimental probability of getting heads? It is 10% higher than the theoretical probability. It is 10% lower than the theoretical probability. It is equal to the theoretical probability for this data. The experimental probability cannot be concluded from the data in the table.
Michele_Laino
  • Michele_Laino
the theoretical probability is 1/2 since we have two possible outcomes, namely a head and a tail and we have one favorable outcome, namely a head, so probability = 1/2= 0.5 The experimental probability to get a head is 90/150=...
Michele_Laino
  • Michele_Laino
since we have 90 favorable cases and 150 possible cases or outcomes
anonymous
  • anonymous
0.6
anonymous
  • anonymous
SO IT 10% lower than the theoretical probability
Michele_Laino
  • Michele_Laino
now the difference between the experimental probability and the theoretical probability is: 0.6-0.5=...
anonymous
  • anonymous
so its 10% higher
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
so the answer is A
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
yay!!!! last question sorry
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
The number of marbles of different colors stored in a hat is listed below: 8 red marbles 10 green marbles 6 blue marbles Without looking in the hat, Tessa takes out a marble at random. She replaces the marble and then takes out another marble from the hat. What is the probability that Tessa takes out a blue marble in both draws? fraction 1 over 16 fraction 1 over 12 fraction 1 over 4 fraction 1 over 2
Michele_Laino
  • Michele_Laino
here we have 8+10+6= 24 possible outcomes and we have 6 favorable outcomes so the probability associated to a single extraction is: p= 6/24 please simplify that fraction
anonymous
  • anonymous
0.25
Michele_Laino
  • Michele_Laino
we can divide both numerator and denominator by 6, wha do you get?
anonymous
  • anonymous
divide what?
anonymous
  • anonymous
oh 1/4
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
so is that the answer @Michele_Laino
Michele_Laino
  • Michele_Laino
now, the probability associated to the second extraction is the same, namely 1/4. So the requested probability is given by the product between the above probabilities, namely p= (1/4)*(1/4)=...?
Michele_Laino
  • Michele_Laino
what is: \[\Large P = \frac{1}{4} \times \frac{1}{4} = ...?\]
anonymous
  • anonymous
Dang it
anonymous
  • anonymous
i just submitted my test
anonymous
  • anonymous
and i got it wrong
anonymous
  • anonymous
because i forgot to do the last part
anonymous
  • anonymous
):
anonymous
  • anonymous
but i got a 87% percent on my test
Michele_Laino
  • Michele_Laino
so have you passed it?
anonymous
  • anonymous
yes i did
Michele_Laino
  • Michele_Laino
CONGRATULATIONS!!!!!
anonymous
  • anonymous
im finished with math until next year yay!!!!!!!!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
Im going to be in 8th grade
Michele_Laino
  • Michele_Laino
ok!! :)
anonymous
  • anonymous
but Atleast i got a high B on my test !
anonymous
  • anonymous
and i have a A in Math!
Michele_Laino
  • Michele_Laino
good job!!
anonymous
  • anonymous
Thank you so much you really helped me undertand what i was doing. But i will be coming back just in case i need help lol!
Michele_Laino
  • Michele_Laino
ok! :)
Jack1
  • Jack1
hi marc just if u could, please only post 1x question per post cheers

Looking for something else?

Not the answer you are looking for? Search for more explanations.