Answer:
we can conclude that there is no significant evidence to conclude that the mean score in 2010 differs from the mean score in 2009.
Step-by-step explanation:
H0 : μ = 582
H1 : μ < 582
Test statistic :
T = (xbar - μ) ÷ σ/√n
Xbar = 515 ; n = 20 ; σ = 120
T = (515 - 582) ÷ 120/√20
T = -67 / 26.832815
T = 2.50
Pvalue at t score = 2.50 ; df = 19 is 0.0187
At α = 0.0187
Pvalue > α ; Hence, we fail to reject the Null
Hence, we can conclude that there is no significant evidence to conclude that the mean score in 2010 differs from the mean score in 2009.
5. Positive
6. Negative
7. Negative
8. Positive
9. No correlation
10. Positive
11. No correlation
12. Positive
C if its a question on plato
Answer:
ok!!!and no one will be Interested seeing that
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.
