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
A whole number has a denominator of 1
Answer:
-4
Step-by-step explanation:
-4 x 13 = -52
Hope that helps :)
Answer:
<h2>x² = -3</h2>
Step-by-step explanation:
In algebra, the goal is always to isolate the variable, so its value can be determined.
<h3>Step 1: Subtract 21</h3>
7x² = -21
<h3>Step 2: Divide by 7</h3>
x² = -3
<h3>Step 3: Check</h3>
7(-3) + 21 = 0
0 = 0 ✔
<h3>Step 4: Answer</h3>
x² = -3
I'm always happy to help :)
Answer:
the answer is slope=94, y-intercept = 2
Step-by-step explanation:
i took the test