65
1. you work in the parentheses and multiply 2.8 by 5 which is 14
2. you divide 14 by 7 which is 2
3. you follow PEMDAS and multiply 26 by what is in the parentheses (2) which is 52
4. you add 13 with 52 which is 65
hope this helped!!
We have given the table of number of male and female contestants who did and did not win prize
The probability that a randomly selected contestant won prize given that contestant was female is
P(contestant won prize / Contestant was female)
Here we will use conditional probability formula
P(A/B) = 
Let Event A = selected contestant won prize and
event B = selected contestant is famale
Then numerator entity will
P(A and B) = P(Contestant won prize and Contestant is female)
= Number of female contestant who won prize / Total number of contestant
= 3 /(4+9+3+10)
= 3 / 26
P(A and B) = 0.1153
P(B) = P(contestant is female )
= Number of female contestant / Total number of contestants
= (3+10) / 26
P(B) = 0.5
Now P(A / B) = 
= 0.1153 / 0.5
P(A / B) = 0.2306
The probability that randomly selected contestant won prize given that contestant is female is 0.2306
Converting probability into percentage 23.06%
The percentage that randomly selected contestant won prize given that contestant is female is 23%
2 beginner, 6 intermediate, 3 advanced...total of 11
P(advanced) = 3/11
replaced
P(beginner) = 2/11
P(both) = 3/11 * 2/11 = 6/121 <=
Answer:
Depends on Howe much you can drink. It varies.
Step-by-step explanation:
Answer:
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
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
90% 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 = 1.7291
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have
. So



The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.