The cost can be optimized by using a Linear Programming given the linear constraint system
- To minimize the cost, the biologist should use <u>60 samples of Type I</u> bacteria and <u>0 samples of Type II</u> bacteria
Reason:
Let <em>X</em> represent Type 1 bacteria, and let <em>Y</em>, represent Type II bacteria, we have;
The constraints are;
4·X + 3·Y ≥ 240
20 ≤ X ≤ 60
Y ≤ 70
P = 5·X + 7·Y
Solving the inequality gives;
4·X + 3·Y ≥ 240
(Equation for the inequality graphs)
The boundary of the feasible region are;
(20, 70)
(20, 53.
)
(60, 0)
(60, 70)
The cost are ;
![\begin{array}{|c|c|c|}X&Y&P= 5\times X + 7 \times Y\\20&70&590\\20&53.\overline 3&473.\overline 3\\60&0&300\\60&70&790\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7Cc%7C%7DX%26Y%26P%3D%205%5Ctimes%20X%20%2B%207%20%5Ctimes%20Y%5C%5C20%2670%26590%5C%5C20%2653.%5Coverline%203%26473.%5Coverline%203%5C%5C60%260%26300%5C%5C60%2670%26790%5Cend%7Barray%7D%5Cright%5D)
- Therefore, the minimum cost of $300 is obtained by using <u>60 samples of Type I</u> and <u>0 samples of Type II</u>
Learn more here:
brainly.com/question/17646656
The ratio the point P partition the directed line segment from M to N is 9:2
<u>Step-by-step explanation:</u>
Let the total distance between M and N be 'd'
Point P distance from M = (9/11) d
Point P distance from N = (1 - 9/11) d
= (2/11) d
The ratio of the distance = (9/11) d / (2/11) d
= 9/2
= 9:2
The ratio the point P partition the directed line segment from M to N is 9:2
1 plus 1 is 100, welcome <3
The greatest common factor is 28.
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, 56
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 28.