In the experiment of rolling two fair dice successively, what is the probability that a sum of 5 appears before a sum of 7? Hint: See Exercise 36. Let S be the sample space of a repeatable experiment. Let A and B be mutually exclusive events of S. Prove that, in independent trials of this experiment, the event A occurs before the event B with probability P (A)/[P (A) + P (B)].
We draw cards, one at a time, at random and successively from an ordinary deck of 52 cards with replacement. What is the probability that an ace appears before a face card?

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!

answer: a) 56% women

b)44% men

step-by-step explanation:

70 years.

step-by-step explanation:

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.