The probability tests to detect the presence of hepatitis B:
P (presence)= 0.97 (true) 0.03 (false)
P (absence)= 0.99 (true) 0.01 (false)
P(infected)=0.0055
P(not infected)= 0.9945
To have an incorrect result:
P[infected and false from P(presence)] + P[not infected and false from P(absence)]
=0.0055*0.03+0.9945*0.01= 0.01011
The answer is<span> 0.01011 </span>
The rule (x, y) ⇒ (x-2, y+3) will translate the way you want.
Answer:
106.8
Step-by-step explanation:
126.16÷106.8
thank you
The answer your looking for is 36
