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Rasek [7]
3 years ago
6

Solve for x. 4(x + 2) = 48 x = 6 1/2 x = 7 1/2 x = 10 x = 14

Mathematics
2 answers:
____ [38]3 years ago
7 0

Answer:

Step-by-step explanation:

4(x + 2) = 48

4x+8=48

4x=40

x=10

alexgriva [62]3 years ago
3 0

[ Answer ]

\boxed{\bold{X \ = \ 10}}

[ Explanation ]

  • Solve For X: 4(x + 2) = 48

--------------------------------

  • Divide Both Sides By 4

\bold{\frac{4(x \ + \ 2)}{4}} = \bold{\frac{48}{4}}

  • Simplify

X + 2 = 12

  • Subtract 2 From Both Sides

X + 2 - 2 = 12 - 2

  • Simplify

X = 10

\boxed{\bold{[] \ Eclipsed \ []}}

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the uniform distribution over the interval 8.5 to 11.5 gallons per minute. Find the probability that between 9.0 gallons and 10.
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Answer: 1/3

Step-by-step explanation:

Here is the complete question:

machine is set to pump cleanser into a process at the rate of 9 gallons per minute. Upon inspection, it is learned that the machine actually pumps cleanser at a rate described by the uniform distribution over the interval 8.5 to 11.5 gallons per minute. Find the probability that between 9.0 gallons and 10.0 gallons are pumped during a randomly selected minute.

The Probability of the above question will be calculated as:

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Step-by-step explanation:

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Read 2 more answers
there has to be at least one operating path. Once a component fails on a path, that path is no longer operating. The reliability
Eva8 [605]

Answer:

R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}

For this case the reliability of the sytem would be given by:

R= \prod_{i=1}^n R_i

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

R = 0.95*0.95*0.5*0.79*0.6= 0.21389

Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}

For this case the reliability of the sytem would be given by:

R= \prod_{i=1}^n R_i

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

R = 0.95*0.95*0.5*0.79*0.6= 0.21389

8 0
3 years ago
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