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.
Answer:
15.766
Step-by-step explanation:
17 menos 1.234 es 15
Answer:
38
Step-by-step explanation:
3 x
{10 +1} +5.
PEMDAS
Parentheses first
3 x
{11} +5.
Then multiply and divide
33+5
38
Answer:
x = 21
Step-by-step explanation:
1. Write the equation
(18 + 2x) ÷ 5 = 12
2. Multiply both sides by 5
18 + 2x = 60
3. Subtract 18 from both sides
2x = 42
4. Solve for x by dividing both sides by 2
x = 21