Look at the ss down below
Answer:
20 inches²
Step-by-step explanation:
perimeter = 8 + 3 + 5 + 4
= 20 inches²
Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.96](https://tex.z-dn.net/?f=z%20%3D%201.96)
Now, find the margin of error M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when
. So
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.3 = 1.96*\frac{4.103}{\sqrt{n}}](https://tex.z-dn.net/?f=0.3%20%3D%201.96%2A%5Cfrac%7B4.103%7D%7B%5Csqrt%7Bn%7D%7D)
![0.3\sqrt{n} = 1.96*4.103](https://tex.z-dn.net/?f=0.3%5Csqrt%7Bn%7D%20%3D%201.96%2A4.103)
![\sqrt{n} = \frac{1.96*4.103}{0.3}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A4.103%7D%7B0.3%7D)
![(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%2A4.103%7D%7B0.3%7D%29%5E%7B2%7D)
![n = 718.57](https://tex.z-dn.net/?f=n%20%3D%20718.57)
Rouding up
We need a sample size of at least 719
Answer:
d^2 = 30^2 + 12^2
e^2 = d^2 + 8^2
e^2 = 30^2 + 12^2 + 8^2
e = √(30^2 + 12^2 + 8^2) = 33.3 ft
A=Annual amount=2000
i=annual interest=0.0205
n=number of years=3
Present value
=A((1+i)^n-1)/(i(1+i)^n)
=2000(1.0205^3-1)/(.0205(1.0205^3))
=5762.15