A=12
120% of a =80% of b
120%a = 80%b
120a/100 = 80b/100
Divide through by 100
120a = 80b
Divide through by 80
120a/80 = b
But a=12
120x 12/80 = b
18 = b
b = 18

a+b = 12 + 18 = 30

8 0

3 years ago

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;

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

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

$\mathbf{{|E|}=1 }$ ; $\mathbf{|R|= 2^{10}}$

$\mathbf{P(E) = \dfrac{1}{2^{10}}}$

4 0

3 years ago

Many poker games start with each player being dealt five cards, called the player's "hand". Remember, there are 52 cards in a st

jek_recluse [69]

Answer:

The number of different possible poker hands is found by counting the number of ways that 5 cards can be selected from 52 cards, where the order is not important. It is a combination, so we use c n r

Step-by-step explanation:

The possible number of the poker hands is probably . C 52 5= 52!/5!x47!= 2598960

5 0

3 years ago