Answer:
Correct option: (a) 0.1452
Step-by-step explanation:
The new test designed for detecting TB is being analysed.
Denote the events as follows:
<em>D</em> = a person has the disease
<em>X</em> = the test is positive.
The information provided is:
Compute the probability that a person does not have the disease as follows:
The probability of a person not having the disease is 0.12.
Compute the probability that a randomly selected person is tested negative but does have the disease as follows:
![P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%29%3DP%28X%5E%7Bc%7D%7CD%29P%28D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%20P%28D%29%5C%5C%3D%5B1-0.97%5D%5Ctimes%200.88%5C%5C%3D0.03%5Ctimes%200.88%5C%5C%3D0.0264)
Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:
![P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%5E%7Bc%7D%29%3DP%28X%5E%7Bc%7D%7CD%5E%7Bc%7D%29P%28D%5E%7Bc%7D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%7B1-%20P%28D%29%5D%5C%5C%3D0.99%5Ctimes%200.12%5C%5C%3D0.1188)
Compute the probability that a randomly selected person is tested negative as follows:
Thus, the probability of the test indicating that the person does not have the disease is 0.1452.
Step-by-step explanation:
13.8564064606
hope it helps
Four is greater than y, and -1 is less than y
Answer:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z=1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).
Answer:
sinΘ = 
Step-by-step explanation:
sinΘ = 
= 
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