Let the lengths of pregnancies be X
X follows normal distribution with mean 268 and standard deviation 15 days
z=(X-269)/15
a. P(X>308)
z=(308-269)/15=2.6
thus:
P(X>308)=P(z>2.6)
=1-0.995
=0.005
b] Given that if the length of pregnancy is in lowest is 44%, then the baby is premature. We need to find the length that separates the premature babies from those who are not premature.
P(X<x)=0.44
P(Z<z)=0.44
z=-0.15
thus the value of x will be found as follows:
-0.05=(x-269)/15
-0.05(15)=x-269
-0.75=x-269
x=-0.75+269
x=268.78
The length that separates premature babies from those who are not premature is 268.78 days
Answer:
12 years
Step-by-step explanation:
20-8=12
Micheal's babysitter is 12 years older than Micheal.
Answer:
pi/2, 330 degres,5pi/3, 7pi/6, 2pi/3 5pi/3, 7pi/6, 2pi/3, pi/2, 330
this the correct answer hope this helps you !!
Answer:
Step 1: Identify the values for b and c. In this example, b=6 and c=8.
Step 2: Find two numbers that ADD to b and MULTIPLY to c. This step can take a little bit of trial-and-error. ...
Step 3: Use the numbers you picked to write out the factors and check
Step-by-step explanation:
thank me later
Answer:
Your answer would be D, the last one
