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3 years ago

5l of 70% alcohol solution was mixed with 9l of pure water what is the concentration of the mixture

Mathematics

1 answer:

ElenaW [278]3 years ago

6 0

The new mixture has a total volume of 5 L + 9 L = 14 L.

70% of the 5 L alcohol solution is alcohol, so that it contributed 3.5 L of alcohol to the mixture. The pure water contributed no alcohol. So in the mixture, the concentration of alcohol is

3.5 L / 14 L = 0.25, or 25%

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A bank in the Bay area is considering a training program for its staff. The probability that a new training program will increas

WITCHER [35]

Answer:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

See explanation below.

Step-by-step explanation:

For this case we define first some notation:

A= A new training program will increase customer satisfaction ratings

B= The training program can be kept within the original budget allocation

And for these two events we have defined the following probabilities

$P(A) = 0.8, P(B) = 0.2$

We are assuming that the two events are independent so then we have the following propert:

$P(A \cap B ) = P(A) * P(B)$

And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:

$P(B' \cup A')$

And using the De Morgan laws we know that:

$(A \cap B)' = A' \cup B'$

So then we can write the probability like this:

$P(B' \cup A') = P((A \cap B)')$

And using the complement rule we can do this:

$P(B' \cup A') = P((A \cap B)')= 1-P(A \cap B)$

Since A and B are independent we have:

$P(A \cap B )=P(A)*P(B) =(0.8*0.4) =0.32$

And then our final answer would be:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

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