Answer:
Step-by-step explanation:
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
Answer: 0.4 pounds of the trail mix is oats.
To find the answer, you can set up a proportion. We can use the percent proportion below.
part/whole = percent/100
x/2 = 20/100
To solve this you cross multiply and divide.
100x = 40
x = 0.4
62.5 is your answer in decimal
