The probability according to the given scenario is:
(a) <em>0.0395</em>
(b) <em>0.95</em>
(c) <em>0.076</em>
The given values are:
(a)
The probability that the b/w has tuberculosis as well as having (+) test will be:
→ 
By substituting the values, we get
→ 
→ 
(b)
The probability that a person does not have tuberculosis will be:
→ 
→ 
→ 
(c)
The person does not have tuberculosis as well as have a (+) test, the probability will be:
→ 
→ 
→ 
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brainly.com/question/14816227
Answer:
Z, Y, X,
W, V,
U, T, S, R, Q, P, O, N,
M, L, K,
J, I, H,
G, F, E, D,
C, B, and A;
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that
Group Group One Group Two
Mean 26.00 23.00
SD 2.00 4.00
SEM 0.40 1.21
N 25 11
where group I represents female servers and group II male servers.
We have to calculate confidence interval for 90% for difference in means
The mean of Group One minus Group Two equals 3.00
df = 34
standard error of difference = 0.993
t critical = 2.034 for 90% df 34
Hence confid. interval at 90%
=Mean diff ±2.034 * std error of diff
= (0.98, 5.02)
<span>150 degrees.
Let's assume the center camera is pointed to at an angle of 0 degrees. Since it has a coverage of 60 degrees, then it will cover the angles from -30 to +30 degrees. Now we'll continue to use the +/- 30 degree coverage for the other two cameras. The second camera is aimed at 45 degrees, so it's range of coverage is 15 degrees to 75 degrees (45 +/- 30). Notice that the range from 15 degrees to 30 degrees is covered by 2 cameras. Now the 3rd camera is pointed at -45 degrees, so its coverage is from -15 degrees to -75 degrees. It also has an overlap with the 1st camera from -15 to -30 degrees.
The total coverage of all three cameras ranges from -75 degrees to 75 degrees. That means that an arc of 150 degrees in total is covered by all three cameras.</span>
The answer to your question is
No
