The expected length of code for one encoded symbol is
where is the probability of picking the letter , and is the length of code needed to encode . is given to us, and we have
so that we expect a contribution of
bits to the code per encoded letter. For a string of length , we would then expect .
By definition of variance, we have
For a string consisting of one letter, we have
so that the variance for the length such a string is
"squared" bits per encoded letter. For a string of length , we would get .
I would say d that what i got.
The constant of proportionality should be 45 miles per hour, and the car should travel 135 miles in three hours.
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%
Answer:
because i learned it in class 2 years ago
Step-by-step explanation:
b. Associative Property of Multiplication