Answer:
-2a^4+8a^3-6a^2
Step-by-step explanation:
2a^2(a-1)(3-a)
2a^2(-a^2 +3a-3+a)
2a^2(-a^2+4a-3)
-2a^4+8a^3-6a^2
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:
it would be a distance of 7 units
Step-by-step explanation:
Fractions are basically divisions.
Ex. 2/4 = 1/2
And whole numbers like 10, get greater when raised to an exponent.
Ex. 10^2 = 100
Answer:
C..... I think it is a answer
