Answer:

Step-by-step explanation:
We are given the following in the question:
Sample size, n = 67
Variance = 3.85
We have to find 80% confidence interval for the population variance of the weights.
Degree of freedom = 67 - 1 = 66
Level of significance = 0.2
Chi square critical value for lower tail =

Chi square critical value for upper tail =

80% confidence interval:

Putting values, we get,

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.
Answer:
C. the initial number of club members, in hundreds
Step-by-step explanation:
The general form of such an expression is ...
(initial value)×(growth factor per period)^(number of periods)
The use of 12 in the exponent suggests that the growth factor of 1.02 is an annual factor. If that is the case, for t months, the membership should be modeled as ...
1.8(1.02^(t/12))
_____
As written, the expression is not an exponential expression. With appropriate parentheses, it might be a good model if t is the number of <em>years</em> (not <em>months</em>), and if the expected growth rate is 2% per month.
1.8(1.02^(12t))
Answer: (7,3)
Step-by-step explanation: