Answer:
yes
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:
1/2
Step-by-step explanation:
4/9 is equal to 4.444 repeating so that would be closest to 1/2
If a 1L IV contains 60meq of kcal and the IV was discontinued after 400ml has infused, then the amount of IV received by the patient is 24meq of kcal.
Calculation for the Amount of IV
It is given that,
1L IV contains 60meq of kcal
⇒ 1000 mL of IV contains 60meq of kcal
⇒ 1 mL of IV will contain 60 / 1000 meq of kcal
As per the question,
The amount of IV infused in the patient = 400 mL
⇒ The amount of IV received by the patient in meq of kcal = 400 × (60 / 1000)
= 4 × 6
= 24 meq of kcal
Hence, the patient receives 24meq of kcal amount of IV.
Learn more about amount here:
brainly.com/question/4567186
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