Answer:
560/9 or around 63 cubes
Step-by-step explanation:
2 2/3 -> 8/3
3 1/3 -> 10/3
2 1/3-> 7/3
7/3 x 8/3= 56/9
56/9 x 10/3 = 560/27
560/27 divided by 1/3
= 560/9
the first 4 would be rounded up tot he nearest dollar and the last one rounded down to the nearest dollar
doing that I estimate $23.00
Answer is C
That "9 minutes" doesn't affect the outcome!
How many pieces of candy are in the bag at the beginning? How many of those are "fruit tart chews?" Write a fraction involving these 2 counts. Remember that Britany immediately eats what she draws from the bag, so the 2nd time around, there are only 19 pieces, not 20. What is the prob. that she will pick a jelly treat on her second draw?
Because these experiments are independent, you can find the joint probability by multiplying the 2 probabilities together. Please show your work.
Answer:
Step-by-step explanation:
2005 AMC 8 Problems/Problem 20
Problem
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
Solution
Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.
See Also
2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
It would be 3,8
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