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Mathematics 13 Online
OpenStudy (anonymous):

WILL GIVE MEDAL In a group of 1500 people, 585 have blood type A, 165 have blood type B, 690 have blood type O, and 60 have blood type AB. What is the probability that a person chosen at random from this group has type B blood?

OpenStudy (anonymous):

165 have the blood type B There are 1500 people selected. This can be represented by: 165/1500 Are you with me so far?

OpenStudy (anonymous):

This can be simplified to 11/100

OpenStudy (anonymous):

SO its 11 %

OpenStudy (anonymous):

Correct :)

OpenStudy (anonymous):

i need help with another :)

OpenStudy (anonymous):

Yeah sure. Ask away.

OpenStudy (anonymous):

A two-way frequency table shows grades for students in college and students in high school.

OpenStudy (anonymous):

Based on this data, are "being in high school" and "GPA above 3.0" independent events?

OpenStudy (anonymous):

1. Yes, P(high school | GPA above 3.0) = P(high school) 2. Yes, P(high school | GPA above 3.0) = P(GPA above 3.0) 3. No, P(high school | GPA above 3.0) ≠ P(high school) 4. No, P(high school | GPA above 3.0) ≠ P(GPA above 3.0)

OpenStudy (anonymous):

I'm a little confused with this one. Sorry

OpenStudy (anonymous):

how about this one .. Carson drives to school the same way each day and there are two independent traffic lights on his trip to school. He knows that there is a 30% chance that he will have to stop at the first light and an 80% chance that he will have to stop at the second light. What is the probability that he will NOT have to stop at either light? 14% 24% 50% 80%

OpenStudy (anonymous):

You should give kirby best response for helping

OpenStudy (anonymous):

As these are consecutive events, the probability of the first even happening is 3/10, therefore it not happening would be 7/10. Which is followed by 8/10 chance of stopping at the second one, it not happening would be 2/10. multiply those two to get 14/100 therefore its 14% :)

OpenStudy (anonymous):

@yayo365

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