Answer:
lemme answer it,I'll put the answer at the comment section
Answer:
<u>There would be 11 books in each stack, to have the same number.</u>
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
Books in stack 1 = 15
Books in stack 2 = 9
Books in stack 3 = 9
2. If the books were rearranged so that the stack had the same number of books how many books would be even?
Total books = Books in stack 1 + Books in stack 2 + Books in stack 3
Total books = 15 + 9 + 9
Total books = 33
Total books/Number of stacks = Number of books per stack to be the same
33/3 = Number of books per stack to be the same
<u>Number of books per stack to be the same = 11</u>
Answer:
10 is real, rational, integer, whole, natural
5/6 is real, rational
sqrt of 24 is real, irrational
0 is real, rational, integer, whole
sqrt 81 is real, rational, integer, whole, natural
Step-by-step explanation:
Classification
Answer:
a) 0.857
b) 0.571
c) 1
Step-by-step explanation:
18 juniors
10 seniors
6 female seniors
10-6 = 4 male seniors
12 junior males
18-12 = 6 junior female
6+6 = 12 female
4+12 = 16 male
A total of 28 students
The possibility of each union of events is obtained by summing the probabilities of the separated events and substracting the intersection. I will abbreviate female by F, junior by J, male by M, senior by S; So we have
P(J U F) = P(J) + P(F) - P(JF) = 18/28+12/28-6/28 = 24/28 = 0.857
P(S U F) = P(S) + P(F) - P(SF) = 10/28 + 12/28 - 6/28 = 16/28 = 0.571
P(J U S) = P(J) + P(S) - P(JS) = 18/28 + 10/28 - 0 = 1
Note that a student cant be Junior and Senior at the same time, so the probability of the combined event is 0. The probability of the union is 1 because every student is either Junior or Senior.
(Extra; but please give me brainliest
