We have the following information:
first urn: 6 green balls and 3 red ones
total: 6 + 3 = 9
second urn: 3 green, 3 white and 3 red
total: 3 + 3 + 3 = 9
third urn: 6 green, 1 white and 2 red
total: 6 + 1 + 2 = 9
a) A green ball is more likely to be obtained, since there are more green balls than red balls, which makes the probability higher.
b) probability of drawing a green, red and white ball.
first urn:
green = 6/9 = 66.66%
red = 3/9 = 33.33%
white = 0/9 = 0%
second urn:
green = 3/9 = 33.33%
red = 3/9 = 33.33%
white = 3/9 = 33.33%
third urn:
green = 6/9 = 66.66%
red = 2/9 = 22.22%
white = 1/9 = 11.11%
c) it would be chosen where the probability of drawing green would be the highest, which means that it would be possible both in the first and in the third ballot box, the probability is equal 66.66%
d) without a green ball, the third ballot box would look like this:
5 green balls, 2 red balls and 1 white ball, with a total of 8.
The probability of drawing would be:
green = 5/8 = 62.5%
red = 2/8 = 25%
white = 1/8 = 12.5%
question 19
equal to: 3's
less than: 2's, 1's
greater than: 0's
question 20
40 gallons, 12 concentrate, 28 water
this is a fraction of 12/28
simplification: (12) divided by 12 / (28) divided by 12
1/2
1 part concentrate 2 parts water per each of the 40 total gallons.
so color in 33% for concentrate and 66% water
I am totally not wrong :D:D:D:D
I would say no it’s not correct, the point is not in the correct place
Answer:
125,44
Step-by-step explanation:
Loss = 112÷100*12
=13,44
Therefore : 112 + 13%
=112 + 13,44
=125,44
Answer: 12 - 4 = 8
Step-by-step explanation:
