The probability that you roll a 4 on a six-sided die is start fraction one over six end fraction . The probability that you flip a coin that lands on heads is start fraction one over two end fraction . The probability that you roll a 4 on a six-sided die and you flip a coin that lands on heads is start fraction one over 12 end fraction . What is the probability of flipping a coin and it landing on heads, given that you rolled a 4 on a six-sided die? Are these two events independent? (4 points) a P(H|4) = start fraction one over two end fraction ; therefore, events are independent because P(H|4) = P(H). b P(H|4) = start fraction one over six end fraction ; therefore, events are dependent because P(H|4) ≠ P(H). c P(H|4) = start fraction one over three end fraction ; therefore, events are dependent because P(H|4) ≠ P(H). d P(H|4) = start fraction one over 12 end fraction ; therefore, events are independent because P(H|4) = P(H).

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation:

yes because they have the same amount of green and blur marbles.

step-by-step explanation:

a) 0,0; 1,1; and 2,2

step-by-step explanation:

these have the exact same coordinates, thus they are the correct function.

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.