Given :
The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by
.
To Find :
Find the time at which the concentration is a maximum. b. Find the maximum concentration.
Solution :
For maximum value of x, K'(x) = 0.
![K'(x) = \dfrac{5(x^2+9)- 5x(2x)}{(x^2+9)^2}=0\\\\5x^2+45-10x^2=0\\\\5x^2 = 45\\\\x = \pm 3](https://tex.z-dn.net/?f=K%27%28x%29%20%3D%20%5Cdfrac%7B5%28x%5E2%2B9%29-%205x%282x%29%7D%7B%28x%5E2%2B9%29%5E2%7D%3D0%5C%5C%5C%5C5x%5E2%2B45-10x%5E2%3D0%5C%5C%5C%5C5x%5E2%20%3D%2045%5C%5C%5C%5Cx%20%3D%20%5Cpm%203)
Since, time cannot be negative, so ignoring x = -3 .
Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6
Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.
Hence, this is the required solution.
Answer:
The probability that Scott will wash is 2.5
Step-by-step explanation:
Given
Let the events be: P = Purple and G = Green
![P = 2](https://tex.z-dn.net/?f=P%20%3D%202)
![G = 3](https://tex.z-dn.net/?f=G%20%3D%203)
Required
The probability of Scott washing the dishes
If Scott washes the dishes, then it means he picks two spoons of the same color handle.
So, we have to calculate the probability of picking the same handle. i.e.
![P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)](https://tex.z-dn.net/?f=P%28Same%29%20%3D%20P%28G_1%5C%20and%5C%20G_2%29%20%2B%20P%28P_1%5C%20and%5C%20P_2%29)
This gives:
![P(G_1\ and\ G_2) = P(G_1) * P(G_2)](https://tex.z-dn.net/?f=P%28G_1%5C%20and%5C%20G_2%29%20%3D%20P%28G_1%29%20%2A%20P%28G_2%29)
![P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}](https://tex.z-dn.net/?f=P%28G_1%5C%20and%5C%20G_2%29%20%3D%20%5Cfrac%7Bn%28G%29%7D%7BTotal%7D%20%2A%20%5Cfrac%7Bn%28G%29-1%7D%7BTotal%20-%201%7D)
![P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}](https://tex.z-dn.net/?f=P%28G_1%5C%20and%5C%20G_2%29%20%3D%20%5Cfrac%7B3%7D%7B5%7D%20%2A%20%5Cfrac%7B3-1%7D%7B5-%201%7D)
![P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}](https://tex.z-dn.net/?f=P%28G_1%5C%20and%5C%20G_2%29%20%3D%20%5Cfrac%7B3%7D%7B5%7D%20%2A%20%5Cfrac%7B2%7D%7B4%7D)
![P(G_1\ and\ G_2) = \frac{3}{10}](https://tex.z-dn.net/?f=P%28G_1%5C%20and%5C%20G_2%29%20%3D%20%5Cfrac%7B3%7D%7B10%7D)
![P(P_1\ and\ P_2) = P(P_1) * P(P_2)](https://tex.z-dn.net/?f=P%28P_1%5C%20and%5C%20P_2%29%20%3D%20P%28P_1%29%20%2A%20P%28P_2%29)
![P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}](https://tex.z-dn.net/?f=P%28P_1%5C%20and%5C%20P_2%29%20%3D%20%5Cfrac%7Bn%28P%29%7D%7BTotal%7D%20%2A%20%5Cfrac%7Bn%28P%29-1%7D%7BTotal%20-%201%7D)
![P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}](https://tex.z-dn.net/?f=P%28P_1%5C%20and%5C%20P_2%29%20%3D%20%5Cfrac%7B2%7D%7B5%7D%20%2A%20%5Cfrac%7B2-1%7D%7B5-%201%7D)
![P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}](https://tex.z-dn.net/?f=P%28P_1%5C%20and%5C%20P_2%29%20%3D%20%5Cfrac%7B2%7D%7B5%7D%20%2A%20%5Cfrac%7B1%7D%7B4%7D)
![P(P_1\ and\ P_2) = \frac{1}{10}](https://tex.z-dn.net/?f=P%28P_1%5C%20and%5C%20P_2%29%20%3D%20%5Cfrac%7B1%7D%7B10%7D)
<em>Note that: 1 is subtracted because it is a probability without replacement</em>
So, we have:
![P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)](https://tex.z-dn.net/?f=P%28Same%29%20%3D%20P%28G_1%5C%20and%5C%20G_2%29%20%2B%20P%28P_1%5C%20and%5C%20P_2%29)
![P(Same) = \frac{3}{10} + \frac{1}{10}](https://tex.z-dn.net/?f=P%28Same%29%20%3D%20%5Cfrac%7B3%7D%7B10%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D)
![P(Same) = \frac{3+1}{10}](https://tex.z-dn.net/?f=P%28Same%29%20%3D%20%5Cfrac%7B3%2B1%7D%7B10%7D)
![P(Same) = \frac{4}{10}](https://tex.z-dn.net/?f=P%28Same%29%20%3D%20%5Cfrac%7B4%7D%7B10%7D)
![P(Same) = \frac{2}{5}](https://tex.z-dn.net/?f=P%28Same%29%20%3D%20%5Cfrac%7B2%7D%7B5%7D)
The <em>correct answer</em> is:
Independent.
Explanation:
The criteria for binomial experiments are:
A fixed number of trials.;
Each trial is independent of the others.;
There are only two outcomes; and
The probability of each outcome remains constant from trial to trial.
This means the outcomes are constant and independent.
Answer:
A is the only solution.
Step-by-step explanation:
A simple way to solve it is to plug in the x and y values
For A, we plug in 2 for x and 3 for y
3-3=5(2-2)
0=5(0)
0=0
Ordered pair A is a solution
For B, plug in 3 for x and 2 for y
2-3=5(3-2)
-1=5(1)
-1 does not equal 5
Therefore only A is a solution