Answer:
B and C
Step-by-step explanation:
Answer:
<h2>A</h2>
Step-by-step explanation:
<em><u>e5uiitweyutwwsuoye34t</u></em>
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
1.(-3/2,3)
2.(5/2,2)
3.(5,4)
4.(7/2,-1)
5.(9/2,-1)
I gave the midpoints what do you mean by the given endpoints
We know that
<span>1½ cup of cleaner-------> (1*2+1)/2-----> 3/2 cup of cleaner
</span><span>the ratio of cleaner to water=(3/2)/12---> 3/(2*12)--> 3/24--> 1/8
the answer is
1/8</span>
