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Illusion [34]
3 years ago
8

What is the error in the proof? The error is in step 3. The reason should be the quotient property of exponents. The error is in

step 3. You cannot use a property of logarithms to prove that same property. The error is in step 2. The reason should be the quotient property of exponents. The error is in step 5. The student did not convert from exponential to logarithmic form correctly when solving for x and y.
Mathematics
1 answer:
galina1969 [7]3 years ago
5 0

Answer:

The error is in step 3. You cannot use a property of logarithms to prove that same property.

Step-by-step explanation:

Here we  the proof of the quotient rule as

If Logₐx = M and Logₐy = N

Then x = a^M and y = a^N

x ÷ y = a^M ÷  a^N = a^{(M-N)}

Take log of both sides we get

Logₐ(x÷y) = Logₐa^{(M-N)}

Logₐ(x÷y) =M-N logₐa

Logₐ(x÷y) =M-N

∴Logₐ(x÷y) = Logₐx - Logₐy

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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!

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Step-by-step explanation:  

Given that;

if n ⇒ ∞

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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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