Let x = Class A
Let y = Class B
Let z = Class C
Equations from problem:
x - 160 = y
2/3x = z
y + 92 = z
Solving through substitution:
2/3x = y + 92
2/3x = (x - 160) + 92
2/3x = x - 68
-1/3x = -68
3(-1/3x) = 3(-68)
-x = -204
x = 204
Now substitute x in the first equation to get y:
204 - 160 = y
44 = y
44 = Class B
There are 44 folded paper cranes in Class B.
Answer:
y= -2x+2.
Step-by-step explanation:
1) according to the condition interception is '2' (point (0;2));
2) slope is: (2+4)/(0-3)= -2;
3) finally the required equation is: y=-2x+2.
P.S. the suggested option is not the only one.
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
