Answer:
(7.7981, 8.5419). The distribution of percentage elongation can be any distribution with finite mean and variance.
Step-by-step explanation:
We have a large sample size n = 56 research cotton samples. Besides, and
. A 95% large-sample CI for the true average percentage elongation is given by
, i.e.,
or equivalently (7.7981, 8.5419). The distribution of percentage elongation can be any distribution with finite mean and variance because we have a large sample.
Answer:
Speed = 20 mi/h
Explanation:
There are two points A and B
Distance between both the points are define with function -20t + 45 miles
where t represent the number of hours of travelling.
Now firstly we find the initial position. At starting time is 0, so we put t=0 in given function of t.
At t=0;
d₀ = -20(0) + 45 = 0 + 45 = 45 miles
Now we find the distance travell after starting first hour.
Than we put t = 1 in the given function
At t = 1;
d₁ = -20(1) + 45 = -20 + 45 = 25 miles
Difference between d₁ & d₀ is
45 - 25 = 20 miles
We see that in one hour, total distace is covered 20 miles
Now we use time, speed, distance relation
Speed = distance/time
Speed=
Speed = 20 miles/hour
That's the final answer.
I hope it will help you.