Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)-> q-d= 6
q-d+d= 6+d
q= 6+d...(2a)
(2a)-> (1) 0.25q+0.10d=3.95
0.25(6+d)=0.10d= 3.95
1.95+0.25d+0.10d= 3.95
0.35d= 3.95-1.5
0.35d/0.35= 2.45/0.35
d= 7...(3)
(3)->(2) q-d= 6
q-7= 6
q=6+7
q= 13
There are 13 quarters and 7 dimes.
P(picking one defective) = 3/10
P(picking a 2nd defective) = 2/9
P(1 and 2 defective) = 3/10 x 2/9 = 6/90 = 0.066
Second method using combination:
³C₂ / ¹⁰C₂ = 1/15 = 0.066
1/2+1/2=2/2=1 so 1/2 was eaten and the other half was leftovers.
Each box would have exactly 8.5 kilograms each in them. Hope this helps mate =)
Answer:
1. x = 2
2. x = 5
3. x = 8
4. x = 28
5. x = 13
6. x = 11
7. x = 0
8. x = 72
9. x = 6
10. x = 38
Step-by-step explanation:
hope that helps!
1. subtract 6 on both sides
2. add 3 on both sides
3. divide both sides by 2
4. multiply both sides by 2
5. add 8 on both sides
6. subtract 9 on both sides
7. add 5 on both sides
8. multiply both sides by 6
9. divide both sides by 4
10. add 18 on both sides
