Answer:an equivalent answer is 4/14 or 18/63
Step-by-step explanation:
i really dont understand stand the question but there
Answer:
Step-by-step explanation:
We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.
Number of sample, n = 41
Mean, u = 43.1 hours
Standard deviation, s = 5.91 hours
For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
43.1 ± 1.645 × 5.91/√41
= 43.1 ± 1.645 × 0.923
= 43.1 ± 1.52
The lower end of the confidence interval is 43.1 - 1.52 =41.58
The upper end of the confidence interval is 43.1 + 1.52 =44.62
Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62
The answer is 3.
If n^11/n^8=27 then, n^3=27.
The cube root of 27 is 3.
So the answer is 3.
Answer:
It should be 15 weeks
Step-by-step explanation:
if you look at the graph you can see every two weeks it goes down by 100, go to the end of the graph and continue from there. you should get 15 weeks till 300 tickets
