Night 1-
Wren 12
Jenni 4
Night 2-
Wren 16
Jenni 9
Night 3-
Wren 20
Jenni 14
Night 4-
Wren 24
Jenni 19
Night 5-
Wren 28
Jenni 24
Night 6-
Wren 32
Jenni 29
Night 7-
Wren 36
Jenni 34
Night 8-
Wren 40
Jenni 39
Night 9-
Wren 44
Jenni 44
It would be 9 nights.
Graph using (x) and (y), Wren being (x) and Jenni being (y)
Answer:
If it is less than 3, Player 1 earns 3 points.
If not, Player 2 earns 2 points.
Step-by-step explanation:
<u>Player 1</u> :
p(N < 3) = p(N = 1 or N = 2) = 2/5
<u>Player 2</u> :
p(N ≥ 3) = p(N = 3 or N = 4 or N = 5) = 3/5
<u>We notice that</u> :
p(N < 3) × 3 = (2/5) × 3 = 6/5
On the other hand,
p(N ≥ 3) × 2 = (3/5) × 2 = 6/5
since ,the probability player 1 win multiplied by the associated number of points (3)
is equal to
the probability player 2 win multiplied by the associated number of points (2).
Then the game is fair.
Y = X/3-1
dang brainly character minimum
1) The average increase in the level of CO2 emissions per year from years 2 to 4 is:
Average=[f(4)-f(2)]/(4-2)=(29,172.15-26,460)/2=2,712.15/2=1,356.075 metric tons. The first is false.
2) The average increase in the level of CO2 emissions per year from years 6 to 8 is:
Average=[f(8)-f(6)]/(8-6)=(35,458.93-32,162.29)/2=3,296.64/2=1,648.32 metric tons. The second is false.
3) The average increase in the level of CO2 emissions per year from years 4 to 6 is:
Average=[f(6)-f(4)]/(6-4)=(32,162.29-29,172.15)/2=2,990.14/2=1,495.07 metric tons. The third is false.
4) The average increase in the level of CO2 emissions per year from years 8 to 10 is:
Average=[f(10)-f(8)]/(10-8)=(39,093.47-35,458.93)/2=3,634.54/2=1,817.27 metric tons. The fourth is true.
Answer: Fourth option: The average increase in the level of CO2 emissions per year from years 8 to 10 is 1,817.27 metric tons.
