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:
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or

Step-by-step explanation:
Given
![f(x) = \sqrt[3]x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5Dx)
Required
Write a rule for g(x)
See attachment for grid
From the attachment, we have:


We can represent g(x) as:

So, we have:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
For:

![2 = n * \sqrt[3]{-1}](https://tex.z-dn.net/?f=2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D%7B-1%7D)
This gives:

Solve for n


To confirm this value of n, we make use of:

So, we have:
![-2 = n * \sqrt[3]1](https://tex.z-dn.net/?f=-2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D1)
This gives:

Solve for n


Hence:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or:

Answer:
24 times
Step-by-step explanation:
P(5 or 6) = 2/6
2/6 = x/72
cross-multiply:
6x = 144
x = 24
Answer:
79
Step-by-step explanation:
60<xy<100
y -2 =x
if it is prime
y cannot be even
so y could be 1,3,5,7,9
x would then be 9,1,3,5,7
but to be between 60 and 100
x must be 7 or 9
so the number must be 79 or 57
57 is not prime so
the number is 79