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
First change the mixed fraction into an improper fraction
7 1/2 = 15/2
Next, change multiply 2 to both numerator and denominator
15 x 2 = 30
2 x 2 = 4
30/4
Each piece needs 3/4 of an inch. Divide 30 with 3
30/3 = 10
c) 10 pieces is your answer
hope this helps
Answer:
9,8,1,0,-10,-15
Step-by-step explanation:
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Answer:
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Answer:
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Step-by-step explanation:
