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alexdok [17]
3 years ago
9

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

in, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.4% carbon dioxide, determine the subsequent amount in the room at any time.What is the concentration at 10 minutes?
Mathematics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is  an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

3*60 =180sec = 0.04%

1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.

for 0.04% = 6000ft3

   ?          = 2000ft3

              = 2000* 0.04)/6000 =0.0133%

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

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2 years ago
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Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

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P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

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P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

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P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

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3 years ago
the volume of a gas varies inversely as the pressure upon it. if the volume of the gas is 160 cubic centimeters at a pressure of
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<h3>How to calculate the equilibrium?</h3>

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