F(x) = [2x - 1 - (x-1) in (x - 1)] / [ 1 - in(x-1)]^2
domain of f = (1, 1+e) U (1 + e, infinty)
hope this helps
Answer:
(A) 0.0013
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 8, \sigma = 2.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%208%2C%20%5Csigma%20%3D%202.5)
The probability that telephone call selected a random will last more than 15.5 min is most nearly
This is 1 subtracted by the pvalue of Z when X = 15.5.
So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{15.5 - 8}{2.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B15.5%20-%208%7D%7B2.5%7D)
![Z = 3](https://tex.z-dn.net/?f=Z%20%3D%203)
has a pvalue of 0.9987
1 - 0.9987 = 0.0013
So the correct answer is:
(A) 0.0013
Answer:
School Sucks!
Step-by-step explanation:
Half of this stuff you will never use in real life!
Ummmmmmmmmm I am so sorry I do not know this
The average amount of corn produced per acre represented in scientific notation can be found using the below formula:
![The \;average\; amount\; of\; corn\; per\; acre=\frac{Bushels \; of\; corn\; produced\; in\; farm}{Area\; of\; farm\; in\; acres}](https://tex.z-dn.net/?f=%20The%20%5C%3Baverage%5C%3B%20amount%5C%3B%20of%5C%3B%20corn%5C%3B%20per%5C%3B%20acre%3D%5Cfrac%7BBushels%20%5C%3B%20of%5C%3B%20corn%5C%3B%20produced%5C%3B%20in%5C%3B%20farm%7D%7BArea%5C%3B%20of%5C%3B%20farm%5C%3B%20in%5C%3B%20acres%7D%20%20%20)
We are given:
![Area \; of\; FARM\; land= 4.2 \times 10^2 \; acres\\ \\ Amount \; of\; corn\; produced=6.72\; \times \; 10^4 \; bushels](https://tex.z-dn.net/?f=%20Area%20%5C%3B%20of%5C%3B%20FARM%5C%3B%20land%3D%204.2%20%5Ctimes%2010%5E2%20%5C%3B%20acres%5C%5C%20%5C%5C%20Amount%20%5C%3B%20of%5C%3B%20corn%5C%3B%20produced%3D6.72%5C%3B%20%20%5Ctimes%20%5C%3B%2010%5E4%20%5C%3B%20bushels%20)
Substituting the above information in the formula, we get...
![The \; Average\; amount\; of\; corn=\frac{6.72 \times 10^4}{4.2 \times 10^2} \\ \\ \frac{6.72}{4.2} \times \frac{10^4}{10^2}\; =\; 1.6 \times 10^{4-2}\; =\; 1.6 \times 10^2\\ \\](https://tex.z-dn.net/?f=%20The%20%5C%3B%20Average%5C%3B%20amount%5C%3B%20of%5C%3B%20corn%3D%5Cfrac%7B6.72%20%5Ctimes%2010%5E4%7D%7B4.2%20%5Ctimes%2010%5E2%7D%20%5C%5C%20%5C%5C%20%5Cfrac%7B6.72%7D%7B4.2%7D%20%5Ctimes%20%5Cfrac%7B10%5E4%7D%7B10%5E2%7D%5C%3B%20%3D%5C%3B%20%20%201.6%20%5Ctimes%2010%5E%7B4-2%7D%5C%3B%20%3D%5C%3B%201.6%20%5Ctimes%2010%5E2%5C%5C%20%5C%5C%20%20)
Conclusion:
![The \; Average \; amount\; of \; corn \; is\; 1.6 \times 10^2 \; Bushes\; per\; corn](https://tex.z-dn.net/?f=%20The%20%5C%3B%20Average%20%5C%3B%20amount%5C%3B%20of%20%5C%3B%20corn%20%5C%3B%20is%5C%3B%201.6%20%5Ctimes%2010%5E2%20%5C%3B%20Bushes%5C%3B%20per%5C%3B%20corn%20)
The average amount of corn produced per acre represented in scientific notation is option B)
bushels per acre