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iren2701 [21]
2 years ago
6

What is the value of (-5)(-8) A 40 B 45 C -40 D 13

Mathematics
2 answers:
mariarad [96]2 years ago
4 0

Answer: 40

Step-by-step explanation:

Novay_Z [31]2 years ago
4 0
-5 times -8 = 40
That should be right because a negative times a negative equals a positive
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Using your knowledge of inverse operations, what does 56 - 25 - 56 equal?
Ivan

Answer:

-25

Step-by-step explanation:

Using the commutative property of addition, we can rearrange the expression to 56-56-25. 56-56 = 0, so the answer is 0 - 25 = -25.

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Tan (25pi/2) = ___<br><br> A). 1<br> B). 0<br> C). -1<br> D). Undefined
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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
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y2-y1/x2-x1

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Answer:

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