There are 5 nickels, 3 dimes, and 9 pennies.
Let d = The number of dimes
Let n = number of nickels
Let p = The number of pennies
As given, there are two more nickels than dimes and three times as many pennies as dimes:
d+2=n
3d=p
0.10d+0.05n+0.01p=0.64
Plug in the value to find d
[tex]0.10d+0.05(d+2)+0.01(3d)=0.64[/tex]
[tex]0.10d+0.05d+0.1+0.03d=0.64[/tex]
[tex]0.18d=0.64[/tex]
[tex]d=3[/tex]
The value of p is equal to 3d
p=3(3)=9
n=d+2
n=3+2
n=5
Hence, there are 5 nickels, 3 dimes, and 9 pennies.
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