The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a) find the probability of a pregnancy lasting 308 days or longer.
b) if the length of pregnancy is in the lowest 22? %, then the baby is premature.

find the length that separates premature babies from those who are not premature.

a) the probability that a pregnancy will last 308 days or longer is?
b) babies who are born on or before days are considered premature

a) P(X>308) = 0.00383

b) 256.42 days

Step-by-step explanation:

Population mean (μ) = 268 days

Standard deviation (σ) = 15 days

a) P(X>308)?

The z-score for any length of pregnancy 'X' is given by:

The z-score for X=308 is:

A z-score of 2.6667 is equivalent to the 99.617-th percentile of a normal distribution. Thus, the probability of a pregnancy lasting 208 days or longer is:

b) X at which P(X) < 0.22?

At the 22-nd percentile, a normal distribution has an equivalent z-score of -0.772. Therefore, the length of pregnancy, X, that separates premature babies from those who are not premature is:

step-by-step explanation: where's the phrase?

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )