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
Uhm, did you, maybe, type in the problem wrong? The answer to this equation is: 18/5 or 3.6, lol, no worries, though!
Answer:
96 employees
Step-by-step explanation:
Given that the standard deviation = 22.8
The width in the question = 12
We solve for the margin of error E.
E = width / 2
= 12/2 = 6
At 99%
Alpha = 1-0.99
= 0.01
Alpha/2 = 0.01/2 = 0.005
Z0.005 = 2.576
Sample size n
= ((2.576x22.8)/2)²
= 95.8
= 96
The number of employees is 96
Thank you!
Answer:
i think the answer is 1.33cm/s
Step-by-step explanation:
you have to use m=y2-y1/x2-x1
Answer: 7.012 x 10−8 ==> Option 3
Step-by-step explanation:
(3.012 x 10−8) + (4 x 10−8)=
(3.012+4) x 10−8=
(3.012+4.000) x 10−8=7.012 x 10−8 ==> Option 3
