530(1+.144/12)(.06) = 32.18 so even that is less than 35
So the answer is 31.80
The answer is b because the answer is b and the people don’t know how to do it
Answer:
(2*2*1.5*1/3)+(2*1.5*4)= 14
Step-by-step explanation:
To find the area of a pyramid, you use the equation of 1/3 *lwh. When we place the values in there, we get 2*2*1.5*1/3, which is equal to 2. Then we solve the bottom part. Since it is a rectangular prism, we use the equation of lwh. When we insert the values, we get 2*1.5*4, which is equal to 12. Now, we take these two values of 2 and 12 and we combine them, giving us our final answer of 14.
Answer:
$77
Step-by-step explanation:
Let Friday be the variable f.
Let Saturday be the variable s.
Substitute 11 for f in the second equation.
Therefore, Kaitlin collected $77 on Saturday
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
