Answer:
0.7385 = 73.85% probability that it is indeed a sample of copied work.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is

In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Identified as a copy
Event B: Is a copy
Probability of being identified as a copy:
80% of 15%(copy)
100 - 95 = 5% of 100 - 15 = 85%(not a copy). So

Probability of being identified as a copy and being a copy.
80% of 15%. So

What is the probability that it is indeed a sample of copied work?

0.7385 = 73.85% probability that it is indeed a sample of copied work.
The factor of three is six factor of two is to 3 is three factor of two is 22 is four and poster of four is 22 Answer
Answer:
a) 1/800 or 0.00125
b) i) 0.0013
ii) 0.001
c) 60%
Step-by-step explanation:
T = [tan(2×30)+1][2cos(30)-1] ÷ (y²-x²)
T = (tan60 + 1)(2cos30 - 1) ÷ (41² - 9²)
T = (sqrt(3) + 1)(sqrt(3) - 1) ÷ 1600
T = (3-1)/1600
T = 2/1600
T = 1/800
T = 0.00125
Error: 0.002 - 0.00125
0.00075
%error
0.00075/0.00125 × 100
60%