To find the answer we simply have to find 12% of 58,800. So to do that, we can multiply it by .12
58,800 • .12 = 7,056
So $7,056 is earned per year
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:
cosine i think tell me if im wrong
Step-by-step explanation:
Answer:
Step-by-step explanation:
You can solve this by cross multiplying. 20 ounces/$7=x ounces/$17. multiple $17 by 20 ounces and $7 by x ounces. (17x20=7x). 17x20=340, so 340=7x. Divide both sides by 7, and you will get x equals about 48.6.
