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:
Step-by-step explanation:
6 and - 2 are the only two solutions to the required quadratic equation.
So, if the variable is represented by X then (X - 6) and (X + 2) will be the only two factors of the polynomial function.
Therefore, the equation is
(X - 6)(X + 2) = 0
⇒ 
If the leading coefficient of the equation is 3 then we can write the equation as (Answer)
31 I think I’m not sure may be wrong
Answer:
Step-by-step explanation:
The question reads much more complicatedly than the actual equation.
Sakura * x = Nanuto
412 x = 634 Divide by 412
x = 634/412
x = 1.54 times
The closest answer is 3/2 as far. 1.54 is just about 3/2.
