On the first roll, you have a 1/3 probability of rolling a 1 or 2 and thus winning $200.
There's a 2/3 probability of not rolling a 1 or 2 on the first roll. On the second roll, there is again a 1/3 probability of rolling a 1 or 2, and 2/3 probability otherwise, so that there is a 1/3*2/3 = 2/9 probability of getting a 1 or 2 and thus winning $100, and 2/3*2/3 = 4/9 probability of losing.
(a) Let be a random variable representing the winnings from playing the game. It has the probability mass function
(b) Compute the expected value of :
0.6% of 1300
Convert <span>0.6%</span><span> to a decimal.
</span><span>0.006⋅1300
</span>Multiply 0.006<span> by </span>1300<span> to get </span><span>7.8.
</span>7.8
Answer:
The 90% confidence interval is 0.575 to 0.625.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.
In which
Z is the zscore that has a pvalue of .
For this problem, we have that:
90% confidence interval
So , z is the value of Z that has a pvalue of , so .
The lower limit of this interval is:
The upper limit of this interval is:
The 90% confidence interval is 0.575 to 0.625.
Answer:
A
Step-by-step explanation:
5/7 x 9/9 = 45/63
I'm assuming there's supposed to be a slash between the 45 and 63!
Answer:
7/3
Step-by-step explanation:
5 + 2 = 7
8 - 5 = 3
7/3