The confidence interval is from 9.81 to 10.19.
We first find the mean of the data:
<span>(9.8+10.2+10.4+9.8+10.0+10.2+9.6)/7 = 10
Next we find the standard deviation:
</span>σ=√([<span>(9.8-10)^2+(10.2-10)^2+(10.4-10)^2+(9.8-10)^2+(10-10)^2+(10.2-10)^2+(9.6-10)^2]/7) = 0.262
The z-score for 95% confidence is found by
1-0.95 = 0.05; 0.05/2 = 0.025; from the z-table, it is 1.96.
The confidence interval is calculated using
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Answer:
P > 9
Step-by-step explanation:
first, you add 3 to both sides,
2p > P + 9
then subtracts P from both sides,
P > 9
your final answer is P > 9
Hope this helps! :)
Answer:
As the sample size is appropriately large, the margin of error is 0.026
Step-by-step explanation:
Answer: the number of minutes of long distance call that one can make is lesser than or equal to 12 minutes.
Step-by-step explanation:
Let x represent the number of minutes of long distance call that one makes.
The first three minutes of a call cost $2.10. After that, each additional minute or portion of a minute of that call cost $0.45. This means that if x minutes of long distance call is made, the total cost would be
2.10 + 0.45(x - 3)
Therefore, the inequality to find the number of minutes one can call long distance for $6.15 is expressed as
2.10 + 0.45(x - 3) ≤ 6.15
2.10 + 0.45x - 1.35 ≤ 6.15
0.75 + 0.45x ≤ 6.15
0.45x ≤ 6.15 - 0.75
0.45x ≤ 5.4
x ≤ 5.4/0.45
x ≤ 12
I hope this is the answer you want