A chemistry teacher needs to mix a 40%

1 answer:

8 0

Let f and e represent the amount of 40% and 80% solutions used...

We are told the f+e=20 which means that f=20-e and we can say that:

(.4f+.8e)/20=.7, using f=20-e in this equation we see that:

.4(20-e)+.8e=14

8-.4e+.8e=14

8+.4e=14

.4e=6

e=15, and since f=20-e then

f=5

So 15 quarts of 80% solution is mixed with 5 quarts of 40% solution to produce a 20 quart mixture of 70% solution.

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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b

jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .

So; as each place is fitted with either 0 or 1

$\mathbf{|R|= 2^{10}}$

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

$\mathbf{P(E) = \dfrac{|E|}{|R|}}$