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.
Answer: B
Step-by-step explanation: I took the quiz and got it right
Also, if you graph the equations on a graphing calculator, the lines are right on top of each other, so the first equation can't be less than -1, so no solutions
Answer:
c
Step-by-step explanation:
because i got 11
120+80=200 I hope this helped you with your problem
To find the probability of pulling something first take note of the total number and the amount of which you want to select.
100 and 10
Now set them up as a fraction
10/100 and simplify
1/10 or 10%