Answer:
The formula which can be used to describe the sequence is:
⇒ 3rd answer
Step-by-step explanation:
The terms of the sequence are -81, 108, -144, 192, ...
∵ 108 ÷ -81 = ![-\frac{4}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%7D%7B3%7D)
∵ -144 ÷ 108 = ![-\frac{4}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%7D%7B3%7D)
∵ 192 ÷ -144 = ![-\frac{4}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%7D%7B3%7D)
- There is a constant ratio between each two consecutive terms
∴ The sequence is a geometric sequence
The formula of the nth term of the geometric sequence is
, where a is the first term and r is the constant ratio between each two consecutive terms
∵ The first term is -81
∴ a = -81
∵ The constant ratio =
∴ r = ![-\frac{4}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%7D%7B3%7D)
- Substitute then in the formula above
∴ ![a_{n}=-81(\frac{-4}{3})^{n-1}](https://tex.z-dn.net/?f=a_%7Bn%7D%3D-81%28%5Cfrac%7B-4%7D%7B3%7D%29%5E%7Bn-1%7D)
Assume that
= f(x)
∴ n = x
∴ ![f(x)=-81(\frac{-4}{3})^{x-1}](https://tex.z-dn.net/?f=f%28x%29%3D-81%28%5Cfrac%7B-4%7D%7B3%7D%29%5E%7Bx-1%7D)
The formula which can be used to describe the sequence is:
![f(x)=-81(\frac{-4}{3})^{x-1}](https://tex.z-dn.net/?f=f%28x%29%3D-81%28%5Cfrac%7B-4%7D%7B3%7D%29%5E%7Bx-1%7D)
Answer:
We need a sample size of 2,071,800 or higher.
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{1-0.9}{2} = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.9%7D%7B2%7D%20%3D%200.05)
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 = 1.645](https://tex.z-dn.net/?f=z%20%3D%201.645)
Now, we 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.
In this problem:
We need a sample size of n or higher, 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)
![0.04 = 1.645*\frac{35}{\sqrt{n}}](https://tex.z-dn.net/?f=0.04%20%3D%201.645%2A%5Cfrac%7B35%7D%7B%5Csqrt%7Bn%7D%7D)
![0.04\sqrt{n} = 1.645*35](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%201.645%2A35)
![0.04\sqrt{n} = 57.575](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%2057.575)
![\sqrt{n} = 1439.375](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201439.375)
![\sqrt{n}^{2} = (1439.375)^{2}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%5E%7B2%7D%20%3D%20%281439.375%29%5E%7B2%7D)
![n = 2,071,800](https://tex.z-dn.net/?f=n%20%3D%202%2C071%2C800)
We need a sample size of 2,071,800 or higher.
He makes $375 a week. Since we know he makes the same amount each week, you set up the equation 12m = 4500. To solve, just divide 4500 (total amount made) by the 12 (the amount of weeks) and you get $375 per week.
Answer:
He has 45 kids now.
Step-by-step explanation:
Answer:She will have:
N = 200 smaller lots
Step-by-step explanation: N = (total acres of land) / (smaller lots of land)
Substituting the values we have:
N = 10 / (1/20)
Rewriting we have:
N = 10 * 20
N = 200 lots