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
Answer:
210° , 330°
Step-by-step explanation:
using the sides of the 30- 60- 90 triangle
with legs 1,
and hypotenuse 2 , then
= 30° ← related acute angle
since
- 
then angle is in third / fourth quadrants , then required angles are
180° + 30° = 210° ← in third quadrant
360° - 30° = 330° ← in fourth quadrant
Answer:
B. 
Step-by-step explanation:
Let p be number of points Michael needs to score in 2nd game to catch up.
We have been given that Janet scored 200 and 400 points in the first 2 rounds of a computer game. So the total points scored by Janet in two games will be:
points.
We are also told that Michael had scored 250 points in the first round and he want to get the same total score as Janet.
So we can find the number of points Michael needs to score to catch up Janet by subtracting the number of points scored by Michael in 1st game from total number of points scored by Janet in two games. We can represent this information in an equation as:

Therefore, the equation
will help Michael to find the number of points he need to catch up Janet and option B is the correct choice.
Answer:
alright alright give us a second
27x to the 3rd power