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:
No they are not equivalent
Answer:
29,30,31
this is basically like counting just add 1
Answer:
Zero Product Property
Step-by-step explanation:
If you have the following options:
A. Multiplication property
<u>B. Zero product property</u>
C. Identity property
D. Transitive property
Answer:
(0,1) , (2,4) , (4,7)
Explanation:
convert the equation into slope-intercept form (this would be y=3/2x+1)
now plug in random numbers for x and solve for y by multiplying them by 3/2 and adding 1.