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
1500 + 0.45x < 1.70x
1500 < 1.70x - 0.45x
1500 < 1.25x
1500/1.25 < x
1200 < x.....so she would have to sell 1201 cards to make a profit
Answer:
Step-by-step explanation:
ans is -4
Writing an equation is the easiest way to figure this out.
b = boys, g = girlsb = 3/4g35 students = g + b35 = g + 3/4g35 = 7/4g35/7/4 = g20 girls35 - 20 = b or 3/4(35) = b15 boys
