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Artemon [7]
3 years ago
10

Rounded to the nearest tenth, what is the length of LN? sin(20°) = LN/8 (8)sin(20°) = LN

Mathematics
1 answer:
blagie [28]3 years ago
8 0

Answer:

2.7

Step-by-step explanation:

8 * sin(20) = 2.736

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171 rounds to 200, and 727 rounds to 700. 700+200= 900.

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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

8 0
3 years ago
Maryland had $(62xsquared + x - 4) on mondays she spent $(11x squared - 33) on a pair of gloves how much money did she have left
Anton [14]

Answer:

$( 51x^2 + x + 29).

Step-by-step explanation:

Amount she had left = original amount - amount spent on the gloves

=  62x^2 + x - 4 - (11x^2 - 33)   (note we place the amount spent on gloves in parentheses because we have to subtract the whole amount)

Now we distribute the negative over the parentheses:

= 62x^2 + x - 4 - 11x^2 + 33     ( note - 33 becomes -33*-1 = +33)

Now simplifying like terms:

= 51x^2 + x + 29  (answer).

7 0
3 years ago
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