22, i think because 3 times 3= 9, and 9 times 3 minus 5= 22
Answer:
Small candies 
Extra large candies 
Step-by-step explanation:
Let small candies 
Extra large candies 
the number of candies is at least
.

Cost of
small candy 
Cost of
extra large candy 
but she has only
to spend

Solve for

Since number of candies should be integer.
let 
total spend
which is more than
, so this combination is not possible.

She has
more so she can buy
more small candy.
Hence small candy 
extra large candy 
PART 1:
Jeremy gives the correct answer.
The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....
Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value.
PART 2:
The long division method is shown in the picture below
PART 3:
As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct
Let the reduced price be x
If he paid for 6 rounds, he paid 1 round at full price of $19 and 5 rounds at reduced price:
19 + 5x = 59
5x = 59 - 19
5x = 40
x = 40 ÷ 5
x = 8
The reduced price is at $8 per round.
<span>Well, the number is decimal form, is in the thousandths place, so the denominator, would be 1000's:
<em><u>0.275 = 275⁄1000</u></em></span><em><u> </u></em>
<span>Then, you have to Simplify 275⁄1000 In which, you would get: <em><u> 11⁄40</u></em></span><span><em><u> </u></em></span>