Answer:
Probability that a smoker has lung disease = 0.2132
Step-by-step explanation:
Let L = event that % of population having lung disease, P(L) = 0.07
So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93
S = event that person is smoker
% of population that are smokers given they are having lung disease, P(S/L) = 0.90
% of population that are smokers given they are not having lung disease, P(S/L') = 0.25
We know that, conditional probability formula is given by;
P(S/L) = 
= P(S/L) * P(L)
= 0.90 * 0.07 = 0.063
So, = 0.063 .
Now, probability that a smoker has lung disease is given by = P(L/S)
P(L/S) = 
P(S) = P(S/L) * P(L) + P(S/L') * P(L')
= 0.90 * 0.07 + 0.25 * 0.93 = 0.2955
Therefore, P(L/S) = 
= 0.2132
Hence, probability that a smoker has lung disease is 0.2132 .
Answer:
1
Step-by-step explanation:
any number to the power 0 is equal to 1
Hi there!
»»————- ★ ————-««
I believe your answer is:
Option C
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
⸻⸻⸻⸻
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
Answer:
Step-by-step explanation:
how many weeks are they doing this for?
The equation of the required plane can be obtained thus:
-4(x + 1) + 4(y + 3) + 3(z - 1) = 0
-4x - 4 + 4y + 12 + 3z - 3 = 0
4x - 4y - 3z = 5
Let x = 1, y = 2, then 4(1) - 4(2) - 3z = 5
z = (4 - 8 - 5)/3 = -9/3 = -3
Thus, point (1, 2, -3) is a point on the plane.
Let a = (a1, a2, a3) and b = (b1, b2, b3) be vectors parallel to the plane.
Then, -4a1 + 4a2 + 3a3 = 0 and -4b1 + 4b2 + 3b3 = 0
Let a1 = 2, a2 = -1, then a3 = (4(2) - 4(-1))/3 = (8 + 4)/3 = 12/3 = 4 and let b1 = -1 and b2 = 2, then b3 = (4(-1) - 4(2))/3 = (-4 - 8)/3 = -12/3 = -4
Thus a = (2, -1, 4) and b = (-1, 2, -4)
Therefore, the required parametric equation is r(s, t) = s(2, -1, 4) + t(-1, 2, -4) + (1, 2, -3) = (2s, -s, 4s) + (-t, 2t, -4t) + (1, 2, -3) = (2s - t + 1, -s + 2t + 2, 4s - 4t - 3)
