Teo has a combination of quarters, loonies, and toonies in his pocket. He knows he has 9 coins in total and that he has at least one of each
type of coin.
a) hwo many outcomes end with over 8$

b) How many different combinations of coins could Teo use to pay for an item
that costs between $10 and $12, if he uses fewer toonies than loonies?

343

Step-by-step explanation:

There are quarters(25¢), loonies ($1), and toonies ($2)

And total number of coins is 9

If loonies ($1), and toonies ($2) are one, then quarters(25¢) is 7

If quarters(25¢), and toonies ($2) are one, then loonies ($1)is 7

If loonies ($1), and quarters(25¢) are one, then toonies ($2) is 7

So possible number number of combination for an item that cost between $10 and $12 is

= 1^3 * 1^3 * 7^3

= 343

hope this helps, stay safe :)

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

what are you trying find out?

step-by-step explanation: