2: Which equation does not belong with the other three? Explain. Thank you so much in advance.
2 answers:
y = x² + 6
the equation of a line in slope-intercept form is
y = mx + c ( m is the slope and c the y-intercept )
y = 0.5x - 0.2, y = 4x + 3 and y = x +
are all in slope- intercept form and represent equations of straight lines
y = x² + 6 is not in this form and is not a straight line equation.
In fact it is the equation of a parabola
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Answer:
a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!
b) 1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!
c) ∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³
Step-by-step explanation:
Given that;
if n ⇒ ∞
p ⇒ 0
⇒ np = Constant = λ, we can apply poisson approximation
⇒ Here 'p' is small ( p=0.01)
⇒ if (n=large) we can approximate it as prior distribution
⇒ let the number of defective items be d
so p(d) = ((e^-λ) × λ) / d!
NOW
a)
Let there be x number of repairs, So they will repair 20x machines on time. So if the number of defective machine is greater than 20x they can not repair it on time.
λ[n0.01]
p[ d > 20x ] = 1 - [ d ≤ 20x ]
= 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!
b)
Similarly in this case if number of machines d > 80x/3;
Then it can not be repaired in time
p[ d > 80x/3 ]
1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!
c)
n = 300, lets do it for first case i.e;
p [ d > 20x } ≤ 0.01
1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.01
⇒ ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.99
⇒ ∑²⁰ˣ_k=0 (λ^k)/k! = 0.99e^λ
∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³
