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
Solution :
Given data :
The mean length of the steel rod = 29 centimeter (cm)
The standard deviation of a normally distributed lengths of rods = 0.07 centimeter (cm)
a). We are required to find the proportion of rod that have a length of less than 28.9 centimeter (cm).
Therefore, P(x < 28.9) = P(z < (28.9-29) / 0.07)
= P(z < -1.42)
= 0.0778
b). Any rods which is shorter than 
cm or longer than 
cm that re discarded.
Therefore,
P (x < 24.84 or 25.16 < x) = P( -59.42 < z or -54.85)
= 1.052
Answer:
11 because 165÷25=11 so yeah
<em>b) 30</em>
- <em>Step-by-step explanation:</em>
<em>Hi there !</em>
<em>
the side opposite the angle of 30 is </em><em>half</em><em> the hypotenuse</em>
<em>x = 2×15</em>
<em>x = 30</em>
<em>Good luck !</em>
