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anonymous
 one year ago
http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%2020150602%20at%207.37.12%20PM_zpseia5o480.png.html
I need help number 18 and 20
please check number 18 did
P(A and B)=P(A) X P(B)
P(A and B) / P(B)= P(A)
= 0.15/0.6
= 1/4 ?
anonymous
 one year ago
http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%2020150602%20at%207.37.12%20PM_zpseia5o480.png.html I need help number 18 and 20 please check number 18 did P(A and B)=P(A) X P(B) P(A and B) / P(B)= P(A) = 0.15/0.6 = 1/4 ?

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phi
 one year ago
Best ResponseYou've already chosen the best response.3I would say \[ P(BA)= \frac{P(A \cap B)}{P(A)} \] so \[ P(A)= \frac{P(A \cap B)}{P(BA)} \] with P(BA)= 0.6 and P(A \(\cap\) B) =0.15 from which you get P(A)= 0.25

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok,Thanks, can you do number 20?

phi
 one year ago
Best ResponseYou've already chosen the best response.3For Q20 there are 20 marbles , 9 of them red. so 9/20 chance of choosing a red marble on the first pick. If you put the marble back in, then you will have 9/20 chance of picking red on the second pick, and 9/20 on the third. thus \[ \left(\frac{9}{20}\right)^3 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3If you do not replace the marble, then the chance of picking a red marble on the first pick is 9/20, on the second 8/19 and on the third 7/18 thus \[ \left( \frac{9}{20}\right)\left( \frac{8}{19}\right)\left( \frac{7}{18}\right) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3for part A, each time you pick is an independent event (because we put the marble back in the bag). That means each pick should be 9/20 and for multiple independent events, you multiply the individual probability to find the probability of all the events occurring.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks , can you have me some more ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3for event A, you count up the number of tiles that are consonants (the problem is y is sometimes a vowel e.g. fly, by, dry, or as a consonant e.g. you, yell, young, ...) not sure how to handle that.

phi
 one year ago
Best ResponseYou've already chosen the best response.3and of course for B, first count the number of tiles that are vowels.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me know if later you can solve , I need help if you can check my work it correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you check my answer ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3Let's assume y is only a consonant. then there are 42 vowels, 2 blanks, 56 consonants the chance of choosing a consonant on the first draw is 56/100 and the chance of choosing a vowel on the 2nd draw is 42/99

phi
 one year ago
Best ResponseYou've already chosen the best response.3and we multiply (56/100)*(42/99)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%2020150602%20at%205.56.17%20PM_zpstctblc9x.png.html \[\frac{ 1 }{6 }\] \[\frac{ 1 }{ 6 } . of .60 = 10 ?\]

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, and the 4 spot came up 10 times

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, next http://i296.photobucket.com/albums/mm168/ontoonex99/Screen%20Shot%2020150602%20at%205.56.23%20PM_zpscjpwjnoy.png \[\frac{ 1 }{ 5 }\times30?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks,next http://i296.photobucket.com/albums/mm168/ontoonex99/Screen%20Shot%2020150602%20at%205.56.08%20PM_zpszlf9z9b0.png \[\frac{ 179 }{ 2237 }=0.0800\] 0.0800 x 100= 8% ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3I would leave the answer as 0.08

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%2020150602%20at%205.56.29%20PM_zpsfhmscrjx.png.html \[\frac{ 526 }{ 2392 }?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://i296.photobucket.com/albums/mm168/ontoonex99/Screen%20Shot%2020150602%20at%206.19.45%20PM_zpshyvuxfv0.png \[5<30 <80<90\]

phi
 one year ago
Best ResponseYou've already chosen the best response.3your picture of the weather does not match up with 9/12000

phi
 one year ago
Best ResponseYou've already chosen the best response.3for the weather question I think they want you to order A , B, C , D from least to most likely

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, http://i296.photobucket.com/albums/mm168/ontoonex99/Screen%20Shot%2020150602%20at%206.20.56%20PM_zpsldqx0cs1.png \[\frac{ 9 }{ 12000 }?\]

phi
 one year ago
Best ResponseYou've already chosen the best response.39/1200 (not 12000) and expected number of defects is 9/1200 * 15000

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 9 }{ 1200 }\times1500?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let go back to number 22 D<B<A<C ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3chance of rain on Fri is 5%, so chance of no rain is 95% i.e. D has prob= 95%

phi
 one year ago
Best ResponseYou've already chosen the best response.3and they want the order smallest to biggest. D will be the biggest chance.

phi
 one year ago
Best ResponseYou've already chosen the best response.3what numbers do you get for A, B , C and D ? for A and C, read off the number from the chart for B and D, do 100  what is on the chart.

phi
 one year ago
Best ResponseYou've already chosen the best response.3I get A 80, B 70, C 90, D 95

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks you . Take me so long ti fix out , http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%2020150602%20at%207.35.09%20PM_zpse2zctvy9.png.html 3) Dependent 4) independent
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