You would be richer with the 1/2 bucket of dimes... Not only because they are worth twice as much, but mainly because they are smaller and it would take a larger # of them to fill the bucket 1/2 way.
Incorrect: 1 half-dollar, 1 quarter, 2 dimes, 1 nickel, and 4 pennies. Correct: If half dollars are not allowed, 3 quarters, 2 dimes, 4 pennies. 9 coins If they are, 1 half-dollar, 1 quarter, 2 dimes, 4 pennies. 8 coins ---- Fixed by the staff at www.joeswebs.com, making your dream website come alive for the lowest price, guaranteed.
There are two possible solutions: You could have one quarter, two dimes, two nickels and forty-five pennies, or you could have two dimes, eight nickels and forty pennies. An easy way of approaching this problem is to start by imagining that you have fifty pennies. You have the right number of coins, but are fifty cents short. Instead of adding other coins, you replace pennies with them: replacing a penny with a nickel gains four cents, a dime gains nine, and a quarter gains twenty-four. You can't possibly use more than two quarters, so there are few cases to consider there: If you replace two pennies with quarters, you've gained forty-eight cents, so you only need two more; but any further replacement will give you too much. If you use one quarter, you need to make up twenty-six more cents in steps of four or nine; it's easy to see that two of each works. Finally, with no quarters, you need to gain fifty cents using increments of four or nine; this yields the second solution.
Martha Ryan Lowry has written: 'Floral art for America' -- subject(s): Flower arrangement
There are 2 ways of answering it, first is this is usually some form of a "trick question" where the answer is, "a half dollar and a nickel, because the half isn't a nickel". However, a more technical answer would be that you would have one half dollar and one half-dime (half dimes were minted from 1792-1873, they were small 90% silver coins half the weight of a dime, the nickel is much larger and started being minted in 1866 and is 75% copper and 25% nickel)
Ervin Walter Curtis has written: 'The application of decision theory and scaling methods to selection test evaluation' -- subject(s): Decision making, Personnel management, Psychometrics
29 dimes and one nickel or 28 dimes and 3 nickels or 27 dimes and 5 nickels or 26 dimes and 7 nickels or 25 dimes and 9 nickels or 24 dimes and 11 nickels or 23 dimes and 13 nickels or 22 dimes and 15 nickels or 21 dimes and 17 nickels or 20 dimes and 19 nickels or 19 dimes and 21 nickels or 18 dimes and 23 nickels or 17 dimes and 25 nickels or 16 dimes and 27 nickels or 15 dimes and 29 nickels or 14 dimes and 31 nickels or 13 dimes and 33 nickels or 12 dimes and 35 nickels or 11 dimes and 37 nickels or 10 dimes and 39 nickels or 9 dimes and 41 nickels or 8 dimes and 43 nickels or 7 dimes and 45 nickels or 6 dimes and 47 nickels or 5 dimes and 49 nickels or 4 dimes and 51 nickels or 3 dimes and 53 nickels or 2 dimes and 55 nickels or one dime and 57 nickels
8 dimes, 1 nickel 7 dimes, 3 nickels 6 dimes, 5 nickels 5 dimes, 7 nickels 4 dimes, 9 nickels 3 dimes, 11 nickels 2 dimes, 13 nickels 1 dime, 15 nickels 17 nickels
6 dimes, 5 dimes 2 nickels, 4 dimes 4 nickels, 3 dimes 6 nickels, 2 dimes 8 nickels, 1 dime, 10 nickels, and 12 nickels.
The formula is (20-2x) N + (x) D for x = 0 to 10 20 nickels, no dimes 18 nickels, 1 dime 16 nickes, 2 dimes 14 nickels, 3 dimes 12 nickels, 4 dimes 10 nickels, 5 dimes 8 nickels, 6 dimes 6 nickels, 7 dimes 4 nickels, 8 dimes 2 nickels, 9 dimes No nickels, 10 dimes
There are 11 ways. The formula is (20-2x) N + (x) D for x = 0 to 10 20 nickels, no dimes 18 nickels, 1 dime 16 nickes, 2 dimes 14 nickels, 3 dimes 12 nickels, 4 dimes 10 nickels, 5 dimes 8 nickels, 6 dimes 6 nickels, 7 dimes 4 nickels, 8 dimes 2 nickels, 9 dimes No nickels, 10 dimes
There are no dimes in nickels, but 20 dimes have the same purchasing value as 40 nickels have.
There are no dimes in nickels. But 25 dimes have the same monetary value as 50 nickels have.
That would be 19 nickels.
111 quarters, zero dimes, zero nickels 110 quarters, two dimes, one nickel 109 quarters, four dimes, two nickels
A 10 dimes and 30 nickels b 21 dimes and 19 nickels c 19 dimes and 21 nickels d20 dimes and 20 nickels
There must be at least one nickel, to account for the 5 cents of the 75 cents. So, one solution would be 1 nickel and 277 dimes. This would be (277 x 0.10) + 0.05 = 27.70 + 0.05 = 27.75 Another solution would be 1 dime and 553 nickels. This would be (553 x 0.05) + 0.10 = 27.65 + 0.10 = 27.75 You will have an odd number of nickels in an solution. Possible solutions include 1 nickel & 277 dimes 3 nickels & 276 dimes 5 nickels & 275 dimes 7 nickels & 274 dimes 9 nickels & 273 dimes up to 553 nickels and 1 dime
you separate a mixture of nickels and dimes by their weight