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:
You are 89 years old.
Step-by-step explanation:
First I subtracted 322 from 500 which got me 178. Then I finally did 178 divided by 2 which got me 89.
Hope this helps you! :)
Answer:
5 x (4+y)
Step-by-step explanation:
Answer:
The population of the town in the year 1962 will be 1730.
Step-by-step explanation:
The population of Smallville can be found by the equation 
, where P(t) is the population at time t, r represents the rate of growth, and I is the initial population.
Now, if there are 225 residents in 1950 and the population of the town grows at a rate of 17% per year, then the population of the town in the year 1962 will be 
(Approximate)
<span>let n+2=u
so, the equation became= [2/u]-[3/u]=5
=> [(2+3)/u]=5
=> 5/u=5
=> u=5/5=1
thus, u=1
we know u=n+2
so, n+2=1
=> n=1-2=-1
so, n=-1</span>
