Because the probability is uniform (equally likely) between 20 and 100 pounds, the probability distribution is uniform between 80 and 100 pounds, with the value 1/(100-20) = 1/80 chocolates/pound.
Let x = random variable that represents pounds of chocolate consumed.
Therefore
P(x < 60) = (60 - 20)*(1/80) = 1/2.
Answer: The probability is 1/2.
Answer:
The margin of error for this estimate is of 14.79 yards per game.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.093
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.
This means that 
What is the margin of error for this estimate?



The margin of error for this estimate is of 14.79 yards per game.
The initial value is 3 (when x=0), and the multiplier is 2, when x=2. The equation can be written as
.. y = 3*(2^(x/2))
X=3y+4, y=y
x+y=180, 3y+4+y=180, 4y+4=180
4y+4=180
4y=176
y=44
x=3(44)+4, x=132+4, x=136
(I am not a very good artist in this case:)
Answer:
(0,6) or 6
Step-by-step explanation:
To find the y-intercept look at the last variable you have, which in this case is 6. Glad to help :)