Answer:
a) It can be used because np and n(1-p) are both greater than 5.
Step-by-step explanation:
Binomial distribution and approximation to the normal:
The binomial distribution has two parameters:
n, which is the number of trials.
p, which is the probability of a success on a single trial.
If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.
In this question:
So, lets verify the conditions:
np = 201*0.45 = 90.45 > 5
n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5
Since both np and n(1-p) are greater than 5, the approximation can be used.
Answer:
y = 0.35x + 2.5
Step-by-step explanation:
20/35 and 21/35. The common denominator is 35.
Answer:
For 3x^2+4x+4=0
Discriminant= = -32
The solutions are
(-b+√x)/2a= (-2+2√-2)/3
(-b-√x)/2a= (-2-2√-2)/3
For 3x^2+2x+4=0
Discriminant= -44
The solutions
(-b+√x)/2a= (-1+√-11)/3
(-b-√x)/2a= (-1-√-11)/3
For 9x^2-6x+2=0
Discriminant= -36
The solutions
(-b+√x)/2a= (1+√-1)/3
(-b-√x)/2a= (1-√-1)/3
Step-by-step explanation:
Formula for the discriminant = b²-4ac
let the discriminant be = x for the equations
The solution of the equations
= (-b+√x)/2a and = (-b-√x)/2a
For 3x^2+4x+4=0
Discriminant= 4²-4(3)(4)
Discriminant= 16-48
Discriminant= = -32
The solutions
(-b+√x)/2a =( -4+√-32)/6
(-b+√x)/2a= (-4 +4√-2)/6
(-b+√x)/2a= (-2+2√-2)/3
(-b-√x)/2a =( -4-√-32)/6
(-b-√x)/2a= (-4 -4√-2)/6
(-b-√x)/2a= (-2-2√-2)/3
For 3x^2+2x+4=0
Discriminant= 2²-4(3)(4)
Discriminant= 4-48
Discriminant= -44
The solutions
(-b+√x)/2a =( -2+√-44)/6
(-b+√x)/2a= (-2 +2√-11)/6
(-b+√x)/2a= (-1+√-11)/3
(-b-√x)/2a =( -2-√-44)/6
(-b-√x)/2a= (-2 -2√-11)/6
(-b-√x)/2a= (-1-√-11)/3
For 9x^2-6x+2=0
Discriminant= (-6)²-4(9)(2)
Discriminant= 36 -72
Discriminant= -36
The solutions
(-b+√x)/2a =( 6+√-36)/18
(-b+√x)/2a= (6 +6√-1)/18
(-b+√x)/2a= (1+√-1)/3
(-b-√x)/2a =( 6-√-36)/18
(-b-√x)/2a= (6 -6√-1)/18
(-b-√x)/2a= (1-√-1)/3
Answer:
$1171.05
Step-by-step explanation:
1500 × (1-(6 / 100))⁴
