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Olin [163]
3 years ago
11

Find the exact values of the remaining five trigonometric functions of theta

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
7 0

Hypotenuse of the triangle containing < theta =  sqrt (60^2 + 11^2) = sqrt 3721 = 61
H = 61 , O = 60 and A = 11

sin theta = O/H = 60/61, cos theta = A/H = 11/61
csc theta = H/O = 61/60, sec theta = H?A = 61/11 and cot theta = 11/60

It's C
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Answer:

a)  P[A/B] = 0,019     or     P[A/B] = 1,9 %

b)  P[A- /B-] = 0,9996       or    P[A- /B-] = 99,96 %

Step-by-step explanation:

Bayes Theorem :

P[A/B]  =  P(A) * P[B/A] / P(B)

The branches of events are as follows

Condition 1        real infection     1/300        and     not infection  299/300

Then

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When the test is done   (virus present)  0,9 (+)    0,15 (-)

2.-   299/300

When the test is done  ( no virus )   0,15  (+)     0,85 (-)

Then:

P(A) = event person infected          P(B)  =  person test positive

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where   P(A)  = 1/300  =   0,0033   P[B/A] = 0,9    

Then P(A) * P[B/A] =  0,0033*0,9  =  0,00297

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P(B) = 0,0033*0,9 + 0,9966*0,15    ⇒  P(B) = 0,1524

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b) Following sames steps:

P[A- /B-] = (299/300) * 0,85  / (299/300) * 0,85 + (1/300 * 0,1)

P[A- /B-] = 0,8471 /0,8474

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