Answer:
false is the correct answer of it if the answer going wrong sorry
Answer:
The probability that all the six people will test negative for the antibody is 0.9472.
The probability that the test comes back positive for at least one of the six people is 0.0528
Step-by-step explanation:
Consider the provided information.
probability that antibody is present will be effective is 99.1% and not present is 99.1% of the time.
Part (A)What is the probability that the test comes back negative for all six people?
Let P(X)= P(Antibody not present)
We want test comes back negative for all six that means antibody is present for all six. Thus X=0
The probability that all the six people will test negative for the antibody is 0.9472.
Part (B) What is the probability that the test comes back positive for at least one of the six people?
Hence, the probability that the test comes back positive for at least one of the six people is 0.0528
Answer: The 75th term is 86
Step-by-step explanation:
Since the sequence consists of all natural numbers which are neither perfect squares nor perfect cubes, we can write out the next ninety numbers starting from 2, and then ruling out the squares and cubes. We then count to the 75th number to the desired term. This is shown in the attachment below.
We can as well sum the number of squares and cubes in the range with the number of the number of the desired term i.e
Number of squares = 8 ( 4,9,16,25,36,49,64,81)
Number of cubes = 2 (8 and 27; excluding 64 which is as well a square)
Desired term = 75
So, 8+2+75 = 85.
Therefore, the 85th term counting from 2 ( i.e 86) is the 75th term of the sequence.
Answer:
Step-by-step explanation:
levels of organic pollutants with 99.7% accuracy
levels of volatile solvents with 99.92% accuracy
levels of chlorinated compounds with 89.7% accuracy
samples:
60% contaminated with organic pollutants
27% contaminated with volatile solvents
13% contaminated with traces of chlorinated compounds
a) probability that the test will signal?
P = 0.997*0.6 + 0.9992*0.27 + 0.897*0.13 = 0.5982
+ 0.269784 + 0.11661 = 0.984594 = 98.45%
b) If the test signals, the probability that chlorinated compounds are present?
P = 0.897*0.13/0.984594 = 0.11661/0.984594 = 0.1184346035 = 11.84%