Answer:
The 99% (two-sided) confidence interval for the true average echo duration μ is between 0 sec and 1.99 sec.
Step-by-step explanation:
We have the sample standard deviation, so we use the student t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 101 - 1 = 100
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.6259
The margin of error is:
M = T*s = 2.6259*0.45 = 1.18
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.81 - 1.18. Answer in seconds cannot be negative, so we use 0 sec.
The upper end of the interval is the sample mean added to M. So it is 0.81 + 1.18 = 1.99 sec
The 99% (two-sided) confidence interval for the true average echo duration μ is between 0 sec and 1.99 sec.
Answer: Seth is travelling at 63.047mi/hr
Step-by-step explanation: First convert 57.1 seconds to hours.
1 hour is equal to 3600s so multiply 57.1sec by 1hr/3600s to get 0.0158 hr
The formula for speed distance over time. v=d/t
v= 1mi/0.0158hr
v= 63.047mi/hr
<span>0.27 * 60 = Take the integer part only to get the minutes, then multiply the decimal part by 60 again to get the second</span>
Answer:
Step-by-step explanation:
Add the two equations together:
(x -y) +(x +y) = (34) +(212)
2x = 246
x = 123
y = x-34 = 89
Malik has 89 foreign stamps and 123 domestic stamps.
