21.5m
i think that may be the answer!!
hopefully this helped some?!!!
Answer:
Step-by-step explanation:
= 6.5%
Answer: provided in the explanation segment
Step-by-step explanation:
(a). from the question, we can see that since that б is known, we can use standard normal, z.
we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?
⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;
x ± z * б / √m
which is
3.15 ± 1.28 * 0.32/√10
= 3.15 ± 0.1295 = 3.0205 or 3.2795
(b). normal distribution of weight (c) б is known
(c). option (a) and (e) are correct
(d). from the question, let sample size be given as S
this gives';
1.28 * 0.32/√S = 0.15
√S = (1.28 * 0.32) / 0.15 = 2.73
S = 7.4529
cheers i hope this helps
Your answer would be approximately 254.5 if you round since 9^2 times pie you would get such
