D equals 38.2 I think because t equals 9.55. You would add -16t with -4t and get -20t. Which would turn into D=-20t^2+382. Since you can't multiply 2 with -20t, you would divide it by 382 which would be 191. D=-20t+191 with this, you would subtract 191 from +191 which would cancel out and it would turn the problem into -191=-20t. You would divide -20 from itself and -191 and get t=9.55. Then you plug t where it was in the problem and then you get that D=38.2. I hope my answer is right and it helps you.
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:
Step-by-step explanation:
Ok so umm I literally have no idea or I’m too lazy but good luck
Answer:
x^12
Step-by-step explanation:
Answer:
6/10 is the answer
Step-by-step explanation:
hope it helps!!!