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:
avn= -8 + (n-1)(-7)
Step-by-step explanation:
arithmetic sequence formula= avn= av1 + (n-1)d
av1= first number in the sequence
d= common difference
n= the number of the term to find
The common difference is -7 so d=-7 and you plug it into the equation. The first number in the sequence is -8 so av1.
There is no specific n to find so it remains n.
I hope this helps! Let me know if this helps.
"log" standing alone actually means "logarithm to the base 10."
Thus, y = log x <=> y = log x
10
y
Stated another way (inverse functions), x = 10
I believe that it will take one day
Slope of the line =(7- -5) / (-3-1) = 12 / -4 = -3
a. point slope form is y + 5 = -3(x - 1) Answer
b rearrange to slope-intercept form:-
y = -3x + 3 - 5
y = -3x - 2 Answer
