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Romashka-Z-Leto [24]
3 years ago
8

Find the simplified form of each expression

Mathematics
1 answer:
astraxan [27]3 years ago
7 0

11.(y^6)² = y^12

12.(x^-3)^-5 x^6 = x^21

13.(6q^6)^–4 = 1/1296q^24

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I need help with this question.​
Tamiku [17]

Answer:

51x

Step-by-step explanation:

3x^3 - 7x + 3x^3 + 4x

27x - 7x + 27x + 4x

20x + 31x

51x

3 0
3 years ago
Plz help me fast!!!
Snezhnost [94]
72 because right number on the top of the bar is six, and 12•6=72
8 0
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

8 0
3 years ago
The answer!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Step2247 [10]
Yeah The Answer Would Be A=299
6 0
3 years ago
Earth, Mars, Saturn, Uranus and Neptune revolve around the sun. Earth takes 1 year, Mars 2 years, Saturn 30 years, Uranus 84 yea
Anettt [7]

Answer:

  30 years

Step-by-step explanation:

2 is a factor of 30, so the LCM of 2 and 30 is 30.

Saturn and Mars will appear in the same direction in the night sky every 30 years.

3 0
3 years ago
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