Answer:
I would guess its D. Because it is the one that makes the most sense.
Step-by-step explanation:
H and L has to be on the same day with no other application so not A, B, or C because there is L in all of them.
Hope I helped!
Answer:
0 tests
Yes, this procedure is better on the average than testing everyone, it makes it less cumbersome.
Step-by-step explanation:
Given the information:
Let P be the probability that a randomly selected individual has the disease = 0.1. N individuals are randomly selected, thereafter, blood samples of each person would be tested after combining all specimens. Should in case one person has the disease then it yields a positive result and test should be set for each person.
Let Y be number tests
For n = 3 there are two possibilities. If no one has the disease then the value is 1 otherwise the value is 4, here P = 0.1
Therefore, for Y = 1
P(Y-1) = P(no one has disease)
= 0.9³
= 0.729
If Y = 4
P(Y-4) = 1-P(y = 1)
= 1 - 0.729 = 0.271
The expected number of tests using this formular gives
E(Y) = 1×0.729 + 4×0.271
E(Y) = 0
Factor the left side:-
x (19 + r) = -37 + w
x = ( -37 + w) / (19 + r) Answer
Option I
(3x+4)cm =(2x+10)cm
3x-2x=6
x=6
Option II
(3x+4)= (x+12)
3x-x=12-4
2x=8
x=4
Option III
(2x+10)= (x+12)
2x-x=12-10
x=2
