Answer:
B. 70
Step-by-step explanation:
We have the sum given by
Now, we open the summation by substituting the values of 'i',
i.e. = (1+11) + (2+11) + (3+11) +(4+11) +(5+11)
i.e. = 12 + 13 + 14 + 15 + 16
i.e. = 70.
Hence, the sum of is 70.
So, option B is correct.
Answer:
1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128
Step-by-step explanation:
There can't be more than seven $5 bills, because they would be worth $33.
So, let's try all other cases and see which fits the requests:
- If there are six $5 bills, they are worth $30. You need three more $1 bills to reach $33. So, you have a total of 9 bills, which is not what we want.
- If there are five $5 bills, they are worth $25. You need eight more $1 bills to reach $33. So, you have a total of 13 bills, which is not what we want.
- If there are four $5 bills, they are worth $20. You need thirteen more $1 bills to reach $33. So, you have a total of 17 bills, which is not what we want.
- If there are three $5 bills, they are worth $15. You need eighteen more $1 bills to reach $33. So, you have a total of 21 bills, which is not what we want.
- If there are two $5 bills, they are worth $10. You need twenty-three more $1 bills to reach $33. So, you have a total of 25 bills, which is not what we want.
- If there is one $5, it is worth $5. You need twenty-eight more $1 bills to reach $33. So, you have a total of 29 bills, which is not what we want.
So, there's no way you can have 31 bills worth $1 and $5 that are worth $33 in total.
Answer:
4/12
Step-by-step explanation:
1/6 - 2/12
2/6 = 4/12