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
9514 1404 393
Answer:
(c) -3/4
Step-by-step explanation:
Subtracting a positive number moves you to the left on the number line. Subtracting a negative number moves you in the opposite direction, to the right.
Here, we start at -2 1/2 = -5/2, and we move 1 3/4 = 7/4 to the right from there. Each mark on this number line is 1/4 unit, so we move 7 marks. The results is ...
-2 1/2 -(-1 3/4) = -5/2 +7/4
= -10/4 +7/4 = -3/4
The answer would be=22.9n
31%.
The car bought most often by both males and females collectively is the white car. That has a total frequency of 0.31 out of 1 or, 31/100
Answer :
$38 + $2(n) where n is the no. of cars.
Step-by-step explanation:
He earns $38 daily which is fixed and additional $2 wage depends on the no. of cars which means no. of cars will determine how many $2's he earns.
25 cars = $38 + $2 (25) = $88
30 cars = $38 + $2 (30) = $98
35 cars = $38 + $2 (35) = $108
40 cars = $38 + $2 (40) = $118
