<span>Probability = 0.063
Fourth try = 0.0973
Let X be the number of failed attempts at passing the test before the student passes. This
is a negative binomial or geometric variable with x â {0, 1, 2, 3, . . .}, p = P(success) = 0.7
and the number of successes to to observe r = 1. Thus the pmf is nb(x; 1, p) = (1 â’ p)
xp.
The probability P that the student passes on the third try means that there were x = 2
failed attempts or P = nb(2, ; 1, .7) = (.3)2
(.7) = 0.063 . The probability that the student
passes before the third try is that there were two or fewer failed attmpts, so P = P(X ≤
2) = nb(0, ; 1, .7) + nb(1, ; 1, .7) + nb(2, ; 1, .7) = (.3)0
(.7) + (.3)1
(.7) + (.3)2
(.7) = 0.973 .</span>
Answer: 10y
Step-by-step explanation: All variables equal one unless assigned a specific answer. So It is basically 9y+1
9y+1=10y
Final Answer: 10y
Hope this helps! :D
Answer:
e t
Step-by-step explanation:
Short answer: 98
Remark
when you have an average of 71, it's like saying that you took 4 quizzes and got 71 on all of them.
Step One
Find the total point count for the 4 quizzes.
T_4 = 71 * 4 = 284
Step Two
Create an equation that will give you the necessary point count to average 86.
284 + 5*x = 86 * 9
Why did we do this? It is because you have 9 quizzes altogether. All nine must be something that gives a point count of 86 on each quiz. You have to put together 9 such quizzes with a total count of 774.
Step three
Solve the equation.
284 + 5x = 774 Subtract 284 from both sides.
5x = 774 - 284
5x = 490 Divide by 5
x = 490/5
x = 98
You must get an average of 98 on the next 5 quizzes
Answer: 98 <<<<<
