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Yuki888 [10]
3 years ago
14

Factor 3x^3+3x^2+x+1

Mathematics
2 answers:
Nat2105 [25]3 years ago
6 0
We can solve it like:
3x^3+3x^2+x+1= (3x^3+3x^2)+(x+1)=
3x^2(x+1)+(x+1)=
= (x+1)(3x^2+1)

Have a good day
bixtya [17]3 years ago
5 0
(3x³ + 3x²) + (x + 1)
3x²(x + 1) + 1(x + 1)
(3x² + 1)(x + 1)

Factored form: (3x² + 1)(x + 1)


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There are n machines in a factory, each of them has defective rate of 0.01. Some maintainers are hired to help machines working.
frosja888 [35]

Answer:

a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

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Step-by-step explanation:  

Given that;

if n ⇒ ∞

p ⇒ 0

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⇒ 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;

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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

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⇒ ∑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

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