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:
h'(t)= -12
Step-by-step explanation:
h'(t)=d/dt (-6t x 2 + 24 + 6)
h'(t)=d/dt (-12t + 24 + 6)
h'(t)=d/dt (-12t + 30)
h'(t)=d/dt (-12t) + d/dt (30)
h'(t)= -12+0
Answer:
point G : (7,4)
point F : (7,1)
point J : (2,4)
point H: (5,4)
Step-by-steWhen you translate you take each point and move it how many units it asks you to move. In this problen you move each point 7 units to the right and 9 units up to get your answer.
Answer:
Somewhere around 6 or 9
Step-by-step explanation: