<h3>Hello there i hope you are having a good day :) Your question : A bag contains the letters from the words SUMMER VACATION. You randomly choose a letter A, and do not replace it. Then you choose another letter A. What is the probability that both letters are A’s? Write your answer as a percent, with the percent symbol. If needed, round to the nearest tenth. </h3><h3>Firstly the word is SUMMER VACATION there are 14 letters in this word and 2 of them with the letter A So this would be 2/14 and 1/7 Then there is 13 letters because we used one already so it would be 1/13 Then the second A would equal 1/13 and this would be 1/7 and 1/13 you would times both given 1/91 now written that in percent with a percent symbol 1.0989011% So this would be 1.91%</h3><h3>Hopefully that helps you ❤</h3>

7 0

3 years ago

What is the median of the data?<br><br> 11,16,12,7,23,9,5,5

Stells [14]

Answer:

5,5,7,*9,11,*12,16,23

Step-by-step explanation:

Since, the data has an even amount of numbers once, you put your numbers in order you place your pointer fingers from the 1st to last number & continue to move closer. There is an even amount of numbers so you would have to add the 2 numbers up & then divide them by 2.

9+11=20 divided by 2=10

6 0

3 years ago

(Algebra) Can anybody help me with this please and thank you

Fudgin [204]

1. A
2. L
3. Q
4. P
5. X
6. H
7. (-4,1)
8.(1,4)
9.(-4,-3)
10.(0,-4)
11. (-2,-1)
12. (5,-4)

5 0

3 years ago

The average of 11 numbers in a list is 14. If the average of 9 of the numbers in the list is 9, than what is the average of the

liberstina [14]

Total value of 11 numbers
14*11
154

Total value of 9 numbers
9*9
81

Total value of other 2 numbers
154 - 81
73

So average of other 2 numbers
73 / 2
36.5

3 0

3 years ago