In dividing an irregular shape into polygons it is permissible for the polygons to overlap
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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
The reason the "+ C" is not needed in the antiderivative when evaluating a definite integral is; The C's cancel each other out as desired.
<h3>How to represent Integrals?</h3>
Let us say we want to estimate the definite integral;
I =
Now, for any C, f(x) + C is an antiderivative of f′(x).
From fundamental theorem of Calculus, we can say that;
where Ф(x) is any antiderivative of f'(x). Thus, Ф(x) = f(x) + C would not work because the C's will cancel each other.
Read more about Integrals at; brainly.com/question/22008756
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Answer:
The % growth rate is 15.8%.
Step-by-step explanation:
Let the function is given by , where f(w) is the number of a specific product produced after w weeks.
Now, given for w = 0, f(0) = 190 = a
So, the function becomes ............. (1)
Now, it is also give that at w = 1, f(1) = 220.
So, from equation (1) we get,
⇒ b = 1.15789.
Now, b = 1 + r
⇒ 1.15789 = 1 + r
⇒ r = 0.15789
Therefore, the % growth rate is 0.15789 × 100% = 15.8% (Approx.)