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Mathematics 17 Online
OpenStudy (anonymous):

An experiment was performed 100 times. Let A be a particular event. What is the largest possible value which n(A) could have?

OpenStudy (experimentx):

what is n

OpenStudy (experimentx):

idk mean??

OpenStudy (anonymous):

11. A weather report indicates that there is a chance of rain in Lancaster County tomorrow morning. Which is a correct interpretation? A) It will rain for hours of the next 24 hours. B) It will rain on of the area of Lancaster County. C) Of past days when conditions were similar, out of 5 of them had some rain. D) None of the other choices.

OpenStudy (experimentx):

i guess D ... 1) statement doesn't say it will rain for hours 2) chance means chance ... not for sure .. "it will rain" is an statement that i will definitely rain. 3) it doesn't say anything about probability ... thought this is also highly probable answer

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

An 8-sided die is rolled times with the following results recorded: Outcome Frequency 1 2 3 4 5 0 6 7 8 One can conclude from this experiment that it is not possible to roll a 5 using this die. A) True B) False

OpenStudy (anonymous):

OpenStudy (experimentx):

could you post here instead ... i really hate to download docs

OpenStudy (anonymous):

An 8-sided die is rolled 133 times with the following results recorded:

OpenStudy (anonymous):

Outcome Frequency 1 25 2 3 3 13 4 15 5 0 6 15 7 11 8 48

OpenStudy (anonymous):

One can conclude from this experiment that it is not possible to roll a 5 using this die true or false

OpenStudy (experimentx):

from experimental data analysis point of view .. true

OpenStudy (anonymous):

explain

OpenStudy (experimentx):

probability of getting 5 = frequency of 5/total no of expts

OpenStudy (anonymous):

A card is drawn from a standard 52-card deck. Which of the following are true? (i) P(getting a red card or getting a queen) = 7/13 (ii) P(getting a or getting a red ) = 1/13 (iii) P(getting an ace or a spade) = 4/13 (i) and (ii) only (i) and (iii) only (ii) and (iii) only (i), (ii), and (iii)

OpenStudy (experimentx):

why don't you calculate individually??

OpenStudy (experimentx):

P(getting a red card or getting a queen) = 26/52+4/52

OpenStudy (anonymous):

unsure how to set up equation

OpenStudy (experimentx):

P(getting a or getting a red ) = same as above

OpenStudy (anonymous):

okay. thanks

OpenStudy (experimentx):

P(getting an ace or a spade) = 4/52+13/52

OpenStudy (anonymous):

once i have calculated the numbers, then what?

OpenStudy (experimentx):

check which of the statements are true ... isn't that what your question asked?

OpenStudy (experimentx):

i hope you got your answer

OpenStudy (anonymous):

A card is drawn from a standard 52-card deck. Let A be the event "you draw either a 9 or a red 8 ." Compute . (Write your answer as a fraction.)

OpenStudy (experimentx):

4/52+4/52

OpenStudy (experimentx):

oops ... 2/52 for the latter one

OpenStudy (anonymous):

compute when is not P(A)

OpenStudy (anonymous):

do i subtract?

OpenStudy (anonymous):

How many elements are in the sample space of the experiment "toss 2 dice and flip a coin?" 3 18 72 None of A, B, or C

OpenStudy (experimentx):

for not p(A) = 1 - P(a)

OpenStudy (experimentx):

18

OpenStudy (anonymous):

I set the equation as: p(A) = 1 - 6/52 am i going in the right direction?

OpenStudy (experimentx):

A card is drawn from a standard 52-card deck. Let A be the event "you draw either a 9 or a red 8 P(a) = 4/52+2/52 not getting P(A) = 1-P(A)

OpenStudy (anonymous):

which, after calculating...46/52

OpenStudy (anonymous):

23. A box contains five blue, eight green, and three yellow balls. If a ball is selected at random, what is the probability that it is green?

OpenStudy (experimentx):

8/(total marbles)

OpenStudy (anonymous):

How many elements are in the sample space of the experiment "toss 2 dice and flip a coin?" 3 18 72 None of A, B, or C

OpenStudy (experimentx):

18

OpenStudy (experimentx):

try another thread ... i might not be online too long now

OpenStudy (anonymous):

thanks for helping :)

OpenStudy (experimentx):

oops sorry for wrong answer, you answer would be 72 ... just forgot two dices were rolled H, 1, 1 H, 2, 1 ,,,, ,,,,, two places for coin X 6 places for dice 1 X 6 places for dice 2 = 72

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