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Andreyy89
2 years ago
12

-5/7 x 1/5 please help

Mathematics
1 answer:
Molodets [167]2 years ago
3 0

-5/7 x 1/5

-5 x 1 = -5

7 x 5 = 35

-5/35 = -1/7

Happy to help :P

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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!

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

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3 years ago
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alekssr [168]

Answer:

true

Step-by-step explanation:

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3 years ago
After 3 years what is the total amount of an investment of $1500 in an account paying 3.5% interest, compounded quarterly?
musickatia [10]
<span>$2,226.60 try to read it agian so easy for me

</span>
3 0
3 years ago
Can someone help me finish this question?
eduard
If 1 is 65 then:

2=180-65=115
3=65
4=115
5=115
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