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olasank [31]
2 years ago
14

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics
1 answer:
RSB [31]2 years ago
3 0

Answer:

7/8

Step-by-step explanation:

To find: number which represents the expression

Solution:

Integers refers to natural numbers, 0 together with negative of rational numbers.

A rational number is the number of the form  where p and q are integers and  .

So,  represents

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Andrew [12]

Answer:

7) 1/4, 1/2, 5/8, 3/4, 7/8. This is in the right order.

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3 years ago
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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astraxan [27]
Nico because 1 cm equals 10 mm, so Jake’s thumb is only 5.5 cm.
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Select all the solutions to the inequality: -2x>10
Jlenok [28]
The answer:

-10, -8, -5

Explanation:

If you replace x with every number and solve the inequality, the one ones that correctly complete the inequality are -10, -8, and -5.
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