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:
Subtract 10 from both sides:
Divide both sides by 
:
Placing 
on the left:
For example, if the total charge is $27.50 and the distance ridden is 250 mi, we can calculate the additional cost per mile, as such:
Answer:
btw it olny 8 points but it is c
Step-by-step explanation:
Answer: (c - 3) / 2 = x
Step-by-step explanation:
If c represents the cookies, then you remove 3 because she ate 3 of the cookies. Since half of the remainder of the cookies were given to her friend, you must divide the product of c - 3, by 2. The reason for the parenthesis is so that it overrules PEMDAS, as parenthesis goes before division.
Step-by-step explanation:
option c 120
earning (in$) for 8 dog houses = 40
earning (in$) for 1 dog house= 40/8 = 5
earning (in$) for 24 dog houses= 24×5 = 120
