Alright.....so..... well the
Whole Number: 4 id greater than one
Fraction: 4/1
I am not a 100% sure but try another source also<span />
Step-by-step explanation:
Your problem → 5y+5/2 / 25y-20/40y^2-32y
5y+5/2÷25y-20/40y^2-32y
=5⋅y+5/2÷25×y-20/40×y^2-32×y
=5y+5/2÷25×y-20/40×y^2-32y
=5y+5/2×1/25×y-20/40×y^2-32y
=5y+y/10-20/40×y^2-32y
=5y+y/10-y^2/2-32y
=5y×10+y-y^2×5-32y×10 ÷ 10
=50y+y-5y2-320y ÷ 10
= -269y-5y^2 / 10
Answer:
Plass 4 every time
Step-by-step explanation:
Answer:
Species
C
C
B
B
A
Step-by-step explanation:
i read the passage hence is how i got my answers
Answer:
1 / 2
Step-by-step explanation:
- First observe that the fate of the last person is determined the moment either the first or the last seat is selected! This is because the last person will either get the first seat or the last seat. Any other seat will necessarily be taken by the time the last guy gets to 'choose'.
- Since at each choice step, the first or last is equally probable to be taken, the last person will get either the first or last with equal probability: 1/2
- Armed with the key observation, we see that the event that the last person's correct seat is free, is exactly the same as the event that the first person's seat was taken before the last person's seat.
- Well, each person had to make a random choice, was equally likely to choose the first person's seat or the last person's seat - the random chooser exhibits absolutely no preference towards a particular seat. This means that the probability that one seat is taken before the other must be 1/2