PLEASE HELP ME WITH THIS ILL GIVE 100 points , brainliest and follow you ❤️
PLEASE HELP ME

1) There are 26 letters in the alphabet.  Thus, 27 people must be chosen because otherwise there could be 1 person per initial.

2) There are 7 days in a week.  Thus, if the first 7 had their birthdays on every day of the week, the 8th person must share a birthday on the same day of the week as another.

3) A phone number can contain 0,1,2,3,4,5,6,7,8,9.  Thus, if the first 10 people have a phone number that ends with each digit, the 11th must share it, as there are no unique options.

4)  100/2 = 50.  Thus, the minimum each bag could have is 50.

5) 19/3 > 6.  The maximum number you could get with numbers less than seven is 6,6,6 = 18.  18<19.  Thus, you need a number 7 or greater.

6) The worst scenario that could happen here is that you grab 8 German books, 10 French books, and 12 Spanish, Russian and Italian books.  This would be 54 books.  Thus, with the 55th book, you must have gotten 12 of at least one subject.  Thus, the answer is 55 books.

7) This is vague.  If she cannot repeat any of the three jokes in the triplet, then she must have 36 jokes for this to be possible, because 12*3=36

8) Each of the remaining 29 students made less than 12 mistakes.  Even if 2 students made 11,10,9,8,7,6,5,4,3,2,1 mistakes, there would still be multiple students left that would have made 11,10,9,8,7,6,5,4,3,2, or 1 mistake(s), meaning that at least 3 students made the same amount of mistakes.

9) 465/95 = about 4.9.  Thus, each table could have less than 6 seats, so we cannot be sure.

Hope it helps <3

Step-by-step explanation:

4.

Bags contain x and y numbers of marbles.

x + y > 100

The least number that is more than 100 is 101.

If x and y are both less than 51, then you have at most a total of 100 marbles.

To have 101 marbles, you must have one bag with at least 51 marbles, and the other bag with the rest. Neither bag must have at least 50 marbles. You can have 1 and 100, but that is still more than 50 marbles in one bag.

6.

You are looking for 12 books in the same language. Let's say you start by picking all 10 French books and all 8 German books. You are up to 18 books. There are no more French or German books, so you can't get to 12 books in one language with those. Now you pick 11 Spanish books, 11 Russian books, and 11 Italian books. That is another 33 books, and still no 12 books in one language. You already picked 18 books + 33 books = 51 books. The very next book must be a Spanish book or a Russian book or an Italian book, and it will make the 12th book of that language. You need to pick 52 books to be sure you have 12 of the a single language.

52 books

7.

You need enough jokes to have 12 different combinations of 3. This is a case in which order does not matter.

If you have 3 jokes, named A, B, and C, you only have enough for 1 year, since ABC, ACB, BAC, BCA, CAB, and CBA are all the same triple of jokes.

This is a combination problem.

nCk = (n!)/[k!(n - k)!]

We know we need more than 3 jokes. Let's try 4.

n = 4

nCk = (4!)/[3!(4 - 3)!]

nCk = 4

We need 12, so 4 is not enogh.

n = 5

nCk = (5!)/[3!(5 - 3)!]

= 5 * 4/2

= 10

Now enough.

Try n = 6

nCk = (6!)/[3!(6 - 3)!]

= 20

With 6 different jokes, you can make 20 different triplets, so there is no need to repeat any triplet in 12 years.

6 jokes

m=-16

step-by-step explanation:

hmu if ur willing to do a algebra a online class i’ll pay 600 hmu on the gram profitpreme