Answer:
ok
Step-by-step explanation:
Use binomial distribution, with p=0.20, n=20, x=3
P(X=x)=C(n,x)p^x (1-p)^(n-x)
P(X>=3)
=1-(P(X=0)+P(X=1)+P(X=2))
=1-(C(20,0)0.2^0 (0.8)^(20-0)+C(20,1)0.2^1 (0.8)^(20-1)+C(20,2)0.2^2 (0.8)^(20-2))
=1-(0.0115292+0.057646+0.136909)
=1-0.206085
=0.793915
X= 6
Y= 0
Z= -2
CHECK
3x + y -2z =22
3(6) + 0 -2(-2) =22
18 + 0 +4= 22
22=22
x+5y+z =4
6 +5(0)+ -2 =4
6 + 0 +-2 = 4
4=4
x+3z=0
6 + 3(-2) =0
6 -6 = 0
0=0
Answer:
1. No, Joe is not correct
2. 
Step-by-step explanation:
Given: b and c are parallel lines
To find:
1. whether the given statement is correct or not
2. 
Solution:
1.
Sum of two angles that form a linear pair is equal to 
(linear pair)
So, Joe is not correct
2.
If two lines are parallel then alternate interior angles are equal.
As b and c are parallel lines,
(alternate interior angles)
Answer:
0.9995
Step-by-step explanation:
10% = 0.10
1 - 0.10 = 0.9
n = number of light bulbs = 7
we calculate this using binomial distribution.
p(x) = nCx × p^x(1-p)^n-x
our question says at most 4 is defective
= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)
= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026
= 0.9995
we have 0.9995 probability that at most 4 light bulbs are defective.
