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belka [17]
3 years ago
9

Kent multiplies both sides of the equation below by an expression. k+12/k=8 Then he moves all the terms to one side of the equal

sign in the resulting equation. Which equation must he solve now?
k^2-8k-12=0

k^2+8k+12=0

k^2-8k-12=0

k^2+8k+12=0

k^2-8+12=0

k^2-8k+12=0
Mathematics
2 answers:
Kaylis [27]3 years ago
6 0

Answer: k^2-8k+12=0


Explanation:


1) Given equation:

          k+12/k=8


2) Multiplication property of equality:


         k(k+12/k)=8k


3) Distributive property:


         k^2+12k/k=8k


4) Simplification:


         k^2+12=8k


5) Subtraction property of equality:


         k^2-8k+12=0


So, that is the equation that he has to solve.

BartSMP [9]3 years ago
3 0
<span>k+12/k=8
Multiply k to remove the k in one side
</span><span>(k+12/k=8) (k)
</span>Therefore, 
k^2+12k-8k=0

Thus,
The equations he must solved now is k^2-8+12=0
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My name is Ann [436]

Answer:

a. The value of alpha is 3.014 and the value of beta is 12.442

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c. The probability that data transfer time is between 50 and 75 ms is 0.176

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To calculate alpha we would have to use the following formula:

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alpha=1,406.25 /466.56

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To calculate beta we would have to use the following formula:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. E(X)=37.5 and V(X)=(21.6)∧2  

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P(X>50)=1−P(X≤50)

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The probability that data transfer time exceeds 50ms is 0.238

c. E(X)=37.5 and V(X)=(21.6)∧2  

​Therefore, P(50<X<75)=P(X<75)−P(X<50)  

Hence, To calculate the probability that data transfer time is between 50 and 75 ms we use the following formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time is between 50 and 75 ms is 0.176

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