<span>Note D = r(t)
20min = 1/3 hr and 10min = 1/6 hr
(1/3hr)30mph + (1/6hr)3mph = 10.5 mi, the total distance traveled.
total distance / total time = average speed = 10.5/(1/3 + 1/6) = 10.5/(2/6 + 1/6)
10.5mi/.5hr = 21mph
Hope This Helps
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Answer:
The t-critical value for 95% confidence interval is ±2.2621
Step-by-step explanation:
We are given the following information in the question:
Sample size, n = 10
Alpha, α = 0.05
We have to find the value of t-critical at 95% confidence interval.
Degree of freedom = n - 1 = 9
The t-critical value for 95% confidence interval is ±2.2621