We will get the number of possible selections, and then subtract the number less than 25 cents.
We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.
That's 5*4*3 = 60 selections
Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels.
To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2 on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).
If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.
And there is 1 other selection less than 25 cents, 2 dimes and no nickels.
So that's 4+3+1 = 8 selections which we must subtract from the 60.
Answer 60-8 = 52 selections of coins worth 25 cents or more.
I thinks it C, this a difficult question but idk, that my closest guess...
Answer:
-25
Step-by-step explanation:
First he bought 3 tires which total costed 105, then returned 2, so he got $70 back, meaning he had $35. Then he found 2 $20 bills, which raised his total to $75. Then, he bought a phone for $90, so he has -25 dollars left
Answer:
4.875?
convert it into an improper fraction
