Answer:
a) P(identified as containing explosives)=P(actually contains explosives and identified as containing explosives)+P(actually not contains explosives and identified as containing explosives)
=(10/(4*106))*0.95+(1-10/(4*106))*0.005 =0.005002363
hence probability that it actually contains explosives given identified as containing explosives)
=(10/(4*106))*0.95/0.005002363=0.000475
b)
let probability of correctly identifying a bag without explosives be a
hence a =0.99999763 ~ 99.999763%
c)
No as even if that becomes 1 ; proportion of true explosives will always be less than half of total explosives detected,
Answer:
SAMPLE RESPONSE( The diagram should have five columns, one for each 25% and one for the total, because 25% is 1
4
of 100%. It should have 2 rows, 1 for percents and 1 for dollar amounts.
Step-by-step explanation:
Edg2020
The solution for this problem would be:
Given that there is 99.999%.
Let denote n as the network servers and p as the reliability of each server.
So the probability that the network uptime = 1 - (1 - p)^n
Therefore, (1-p) ^n = 0.00001
a. x= log(1-.99999)÷log(1-.97)= 3.2833 is the answer
1-(1-.97)^3= 0.99999 + 0.0001 = 1
b. x = log(1-.99999)÷log(1-.88) = 5.43 is the answer
1-(1-.88)^3= 0.99 + 0.0001 = approx 1
Answer:74.55 beat per minute
Step-by-step explanation:
Given
Heart beat is 52 beat per minute
i.e. 
beat per sec
If heart rate is increases 1.25 % each sec
suppose B is the heart beat at first second therefore heat beat at 2 nd second
For 3 rd sec
similarly for 3 sec
