1.) 2000 2.) 680000 3.) .0000007 4.) .00135 5.) 3062000 6.) 7x10^5 7.) 3.2x10^7 8.)4.5x10^-2 9.)3.46781x10^-5 10.) 1.3x10^18
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
UGH I think I did this before but I totally forgot how to do it. Give me a minute to do some research and remember because I'm about 79% sure I know how to do this!
Answer: b: h=2a/b1+b2
Step-by-step explanation:
Just get only one of the placeholder on one side by itself
and do the same thing to both sides so it stays equal
3x+7=5x-21
add 21 to both sides
3x+28=5x
subtract 3x from obht sides
28=2x
sivide both sides by 2
14=x
