Answer:
The marginal-cost function is;
c'(x)= (20x-1)/(2√(400 + 10x² - x))
Completed question;
The cost of producing x bags of dog food is given by C(x) = 800 + √(400 + 10x² - x) where 0≤x≤55000. Find the marginal-cost function.
The marginal-cost function is c'(x)=
(Use integers or fractions for any numbers in the expression.)
Step-by-step explanation:
The marginal-cost function is c'(x)= = dC(x)/dx
Where;
C(x) = 800 + √(400 + 10x² - x)
c'(x) = d(800 + √(400 + 10x² - x) )/dx
Using function of function rule of differentiation or chain rule;
c'(x) = (20x-1)×(1/2√(400 + 10x² - x))
c'(x) = (20x-1)/(2√(400 + 10x² - x))
The marginal-cost function is;
c'(x)= (20x-1)/(2√(400 + 10x² - x))
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Answer:
the probability is 0.24 or 24%
Step-by-step explanation:
Given
Percentage of time he goes out with his girlfriend = 40%
Percentage of time he goes out to a bar = 60%
If he goes out with his girlfriend, the percentage of times he spends the night at his girlfriend’s apartment = 30%
If he goes out to a bar, percentage of times he gets in a fight and gets thrown in jail = 40%
let probability that the roommate went to the bar = A
As the roommate going to jail is a conditional probability dependent on A happening, the true probability of the roommate is found by multiplying the absolute probability of the roommate going out to a bar (A) by the probability of the roommate going to jail if he goes to a bar
that is Pr(B) =Pr (B n A)
Pr(B) = Pr(B/A) *Pr(A)
This is given as 60% * 40% or 0.6 * 0.4 (probabilities are 1/100 of percentages)
This gives an answer of 0.24 or 24%
Answer:
The number of rainfalls is 
The answer to the second question is no it will not be valid this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid.
Step-by-step explanation:
from the question we are told that
The standard deviation is 
The margin of error is 
Given that the confidence level is 95% then we can evaluate the level of significance as
Next we will obtain the critical value of 
from the normal distribution table , the value is 
Generally the sample size is mathematically represented as
![n = [\frac{Z_{\frac{\alpha }{2} * \sigma }}{ E} ]^2](https://tex.z-dn.net/?f=n%20%20%3D%20%20%5B%5Cfrac%7BZ_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%2A%20%20%5Csigma%20%7D%7D%7B%20E%7D%20%5D%5E2)
substituting values
![n = [\frac{1.96 * 0.5 }{ 0.1} ]^2](https://tex.z-dn.net/?f=n%20%20%3D%20%20%5B%5Cfrac%7B1.96%20%2A%200.5%20%7D%7B%200.1%7D%20%5D%5E2)
The answer to the second question is no the validity is null this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid
Its will be 50 pairs
starting with 0+99=99 1+98=99
2+97=99 and so on
