The answer is -13.
Solution:
= |-4b - 8| + |-1 - b^2| + 2b^3
= |-4(-2) - 8| + |-1 - -2^2| + 2(-2)^3
= |8-8| + |-1+4| + 2(-8)
= |0| + |3| + (-16)
= -13
20 extra Cds is the answer to the problem but all the inequalitys are so much together that I can not read sorry
In order to solve or know the probability of having 2 girls
and 2 boys, assumed that a girl is as likely as a girl at each birth, pascal’s
triangle will be likely used. And we will be referring to the line 4 of pascal’s
triangle, which was 1 4 6 4 1. Then it
will look like this: 1 = 4 girls; 4 = 3 girls & 1 boy; 6 = 2 girls & 2
boys; 4 = 3 boys & 1 girl; 1 = 4 boys. And now for the solution in order to
get the probability of having 2 girls and 2 boys is to divided into the sum of 1+4+6+4+1.
Answer:
OMGGGG I NEED THIS ONEE TOOOO WANNANA WORKKKK ON OT
3(2(5)+4)-5
3(10+4)-5
30+12-5
42-5
37
