Answer:
0.44
Step-by-step explanation:
Given the estimated logistic regression model on risk of having squamous cell carcinoma
-4.84 + 4.6*(SMOKER)
SMOKER = 0 (non-smoker) ; 1 (SMOKER)
What is the predicted probability of a smoker having squamous cell carcinoma?
exp(-4.84 + 4.6*(SMOKER)) / 1 + exp(-4.84 + 4.6*(SMOKER))
SMOKER = 1
exp(-4.84 + 4.6) / 1 + exp(-4.84 + 4.6)
exp^(-0.24) / (1 + exp^(-0.24))
0.7866278 / 1.7866278
= 0.4402863
= 0.44
Answer:
5 mins difference
Step-by-step explanation:
1 3/4 hours = 1 hour and 45mins
50-45=5mins
Answer:
Answer B is the correct answer
Step-by-step explanation: :)
That will be the second one, square root of 34
Answer:
1
Step-by-step explanation:
we have the probability of dyng = 1.3
we have probability of ying =
p0 = p1 = p2 = p3 = 1/3
d = dying

we cross multiply from here
3d = 1+d+d2
3d-d = 1+d2
2d = 1 + d2
(d-1)2 = 0
2d - 2 = 0
2d = 2
divide through by 2
d = 1
therefore there is a a confirmed probability that the ameba from a single cell is eventually going to die out.