If by solving you mean find the roots, all you have to do is plug 0 in for g(x)
0=-3x+1
-1=-3x
-1/-3=-3x/-3
1/3 = x
Simplifying
2x + -3(3 * 0.6x + 2.7)
Multiply 3 * 0.6
2x + -3(1.8x + 2.7)
Reorder the terms:
2x + -3(2.7 + 1.8x)
2x + (2.7 * -3 + 1.8x * -3)
2x + (-8.1 + -5.4x)
Reorder the terms:
-8.1 + 2x + -5.4x
Combine like terms: 2x + -5.4x = -3.4x
-8.1 + -3.4x
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
That is,
Consider X be the length of the pregnancy
Mean and standard deviation of the length of the pregnancy.
Mean 
Standard deviation \sigma =15
For part (a) , to find the probability of a pregnancy lasting 308 days or longer:
That is, to find 
Using normal distribution,



Thus 
So 




Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.
This the answer for part(a): 0.00256
For part(b), to find the length that separates premature babies from those who are not premature.
Given that the length of pregnancy is in the lowest 3%.
The z-value for the lowest of 3% is -1.8808
Then 
This implies 
Thus the babies who are born on or before 238 days are considered to be premature.