Answer:
The ways in which he can make selection = 
Step-by-step explanation:
Student is having a list of
books.
Assuming that the student can't read the same book twice.
The student has
choices to make for an outside reading requirement,for his 1st choice, he had
books to choose from; for his 2nd, he has
books left to choose from therefore the total permutations of his choices are 
So the total ways in which the student can make his selection = 
To solve follow these steps:
3n+2=83+8n
3n=81+8n (subtract 2 on each side)
-5n=81 (Subtract 8n)
n=-81/5 (divide by -5)
~I hope this helps!~
The probability is 0.1429.
The probability that the first person chosen is a boy is 5/7. The probability that the second is a boy is 4/6; the third, 3/5; and the fourth, 2/4. Both the numerator and denominator decrease every time, because you lose a boy to choose from and you lose a student to choose from as well. This gives us:
5/7(4/6)(3/5)(2/4) = 120/840 = 0.1429.
Answer:
D. 8 1/2 ÷ 5/6 = x
Step-by-step explanation:
let x be the number of smaller boards
Dividing the total amount of board by the length of each board gives the number of smaller boards.
The division statement is in the format:
Total ÷ number of groups = amount per group
OR
Total ÷ amount per group = number of groups
Each group is a small board. The amount per group is the amount of feet for each board.
8 1/2 ÷ 5/6 = x
Answer: frist goes to second. Second goes to the third. Thrid goes to the frist.
Step-by-step explanation: