Answer:
The required equation is
.
Step-by-step explanation:
Consider the provided information.
The Initial value of poster = $ 18
After 1 year amount of increase = $ 20.70
With the rate of 15% = 0.15
Let future value is y and the number of years be x.

Now verify this by substituting x=1 in above equation.

Which is true.
Hence, the required equation is
.
Answer:
- 5 = 76
Step-by-step explanation:
= 81
x = 9
Answer:
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that the selection of the random pages will contain at least two errors is 0.2644
Step-by-step explanation:
From the information given:
Let q represent the no of typographical errors.
Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let
be the random variable that follows a Poisson distribution, then mean 
and the mean that the random selection of 50 pages will contain no error is 
∴

Pr(q =0) = 0.368
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that 50 randomly page contains at least 2 errors is computed as follows:
P(X ≥ 2) = 1 - P( X < 2)
P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2
![P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20%5B%20%5Cdfrac%7Be%5E%7B-1%7D%201%5E0%7D%7B0%21%7D%20%2B%5Cdfrac%7Be%5E%7B-1%7D%201%5E1%7D%7B1%21%7D%20%5D)
![P(X \geq 2) = 1 - [0.3678 +0.3678]](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20%5B0.3678%20%2B0.3678%5D)

P(X ≥ 2) = 0.2644
The probability that the selection of the random pages will contain at least two errors is 0.2644
The formula for interval estimate would be: μ = M ± Z(<span>sM</span>)
Where: μ is estimate
M is the mean
Z is the z value
(<span>sM</span>) is the standard
error
μ = M ± Z(<span>sM</span>)
n = 200 rather than 50 (√200 = 2√50)
<span>⇒ ME = (1/2) * 1.32 = .66</span>
<span>Using the formula above, plugging this in will give us: μ
= 19.76 ± .66</span>
<span> = 19.76 ± .66 is
the confidence interval or interval estimate</span>
Answer:
We have to get the average transaction time down to 371 seconds or 6 minutes 11 seconds.
Step-by-step explanation:
The average transaction time for this location is 6 minutes and 52 seconds.



The average transaction time for this location is

The average transaction time is reduced by 10%.
The average transaction time of this location reduced to 371 seconds.
The average transaction time of this location reduced to 6 minutes 11 seconds.