Total coins = 10 ( 4 + 3 + 3 = 10).
The probability of picking a nickel first = 4/10 = 2/5 ( 4 nickels out of 10 total).
The probability of picking a dime second = 3/9 = 1/3 ( 3 dimes out of 9 coins left).
The probability of picking the third coin as a quarter = 3/8 ( 3 quarters out of 8 coins left).
The probability of picking all three in that order = 2/5 x 1/3 x 3/8 = 1/20
Answer:
are you sure this is the right question? none of those have a sum of 23
Step-by-step explanation:
If xy=0 we assume x and y equal 0
so
the zeros are wehre f(x)=0
0=5(2x-5)(5x+4)
set each to zero
5 is not equal 0 so we don't do that
0=2x-5
5=2x
5/2=x
0=5x+4
-4=5x
-4/5=x
zeroes at x=5/2 and -4/5
