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stich3 [128]
3 years ago
12

Subtract 5x^2 – 2x – 5 from – 2x^2 - 5x - 7

Mathematics
2 answers:
Paladinen [302]3 years ago
6 0

Answer:

-7x² -3x - 2

Step-by-step explanation:

(-2x²-5x -7) - (5x² - 2x - 5)

removing parentheses

-2x² -5x - 7 -5x² + 2x + 5

collect like terms

-2x² - 5x² -5x + 2x -7 + 5

= -7x² -3x - 2

user100 [1]3 years ago
3 0

Answer:

-7x^{2} -3x-2

Step-by-step explanation:

(-2x^2-5x-7) - (5x^2-2x-5)\\\\-2x^2 - 5x^2  = -7 x^2\\-5x - -2x = -3x\\-7 - -5 = -2 \\Result: -7x^2-3x-2

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Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
My name is Ann [436]

Answer:

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

b. The probability that data transfer time exceeds 50ms is 0.238

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

Step-by-step explanation:

a. According to the given data we have that the mean and standard deviation of the random variable X are 37.5 ms and 21.6.

Therefore, E(X)=37.5 and V(X)=(21.6)∧2  

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

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

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  

Therefore, P(X>50)=1−P(X≤50)

Hence, To calculate the probability that data transfer time exceeds 50ms we use the following formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

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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The solution is X=0.4

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