Lisa has 24 flowers stickers, 18 animal stickers and 30 star stickers. She wants to arrange all the stickers in to books. Each book has the same number of stickers and the same type of sticker how many stickers can be in each book?

The number of stickers that would be in each book is :

3 animal stickers, 4 flower stickers, 5 star stickers

Step-by-step explanation:

Lisa has 24 flowers stickers, 18 animal stickers and 30 star stickers.

We solve this above question, using the Greatest Common Factor Method

We find the factors of 24, 18 and 30

The factors of 18 are: 1, 2, 3, 6, 9, 18

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30

Then the greatest common factor is 6.

We are told that: Each book has the same number of stickers and the same type of sticker how many stickers can be in each book?

For animal stickers

18 animal stickers

= 18 animal stickers/6

= 3 animal stickers

For flower stickers

24 flowers stickers

= 24 flowers stickers/6

= 4 flower stickers

For animal stickers

18 animal stickers

= 18 animal stickers/6

= 3 animal stickers

For star stickers

30 star stickers.

= 30 star stickers/6

= 5 star stickers

Therefore, the number of stickers that would be in each book is :

3 animal stickers, 4 flower stickers, 5 star stickers

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

there are infinitely many solutions.