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
Answer:
2×12^(1/4)
Step-by-step explanation:
Step 1: Express 192 as a multiple of primes:
192=2×2×2×2×2×2×3= 2^4 × 2^2 × 3
Step 2: We can take the fourth root of 2^4 and get 2 but we cannot take the fourth root of 2^2 or 3.
Step 3: Now we have 2 × fourth root of 2^2 ×3 = 2 × 12^(1/4)
Answer:
N/A
Step-by-step explanation:
What is line AC, was there supposed to be a picture?
Answer:
27
Step-by-step explanation:
54 of 2
= 27
54 divided by 2
= 27
Yes, I believe this is correct.