Answer: the probability that a person who tested positive actually has TB is 0.64
Step-by-step explanation:
lets say
T : has tuberculosis
+ : test results +ve
- : test result is -ve
so given that
P(+ITc) = 0.008
P(-ITc) = 0.15
P(T) = 50/3000 = 1/60
P(Tc) = 1 - P(T) = 1 - 1/60 = 59/60
Now required probability
= P(TI+)
P(T∩+) / P(+)
= P(T) × P(+IT) / [[P(T) × P(+IT)] + [(P(Tc) × P(+ITc)]
= P(T) × {1-P(-IT)] / [[P(T) × [1-P(-IT)] + [(P(Tc) × P(+ITc)]
WE SUBSTITUTE
= { 1/60 × (1-0.15) } / [1/60 × (1 - 0.15)] + [(59/60) × 0.008]
= (0.0167 × 0.85) / [(0.0167 × 0.85) + (0.9833 × 0.008)]
= 0.01417 / 0.02206
= 0.6423 ≈ 0.64
∴ the probability that a person who tested positive actually has TB is 0.64
1.find the least common denominator
2.find the equivalent fractions
3.add or subtract the fractions and add or subtract the whole numbers
4.write your answer in lowest terms
This might help but i tried
Answer:
Let x = the number
A negative number is 42 less than its square ⇒
x = x2 - 42
x2 - x - 42 = 0
(x - 7)(x + 6) = 0
x = 7 or -6
But since it is given that x is negative, the final answer is x = -6.
Answer:
11,968
Step-by-step explanation:
I'm not sure, but I think its 11,968
(Y1-Y2)/(X1-X2)
(2-4)/(-2-3)
-2/-5
2/5
