Answer:
Step-by-step explanation:
If we take out the extra $3, we can group the bills into one each of $5 and $1, for a value of $6. There will be 7 such groups in the remaining $42.
That means there are 7 bills of the $5 denomination, and 3 more than that (10 bills) of the $1 denomination.
There are 7 $5 bills and 10 $1 bills.
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If you want to write an equation, it is usually best to let a variable stand for the most-valuable contributor. Here, we can let x represent then number of $5 bills. Then the value of the cash box is ...
5x +(x+3) = 45
6x = 42 . . . . . . . . subtract 3, collect terms
x = 7 . . . . . . . . . . . there are 7 $5 bills
x+3 = 10 . . . . . . . . there are 10 $1 bills
You may notice that this working parallels the verbal description above. (After we subtract $3, x is the number of $6 groups.)
(85 + 84 + 78 + x) / 4 = 86
(247 + x) / 4 = 86
247 + x = 86 * 4
247 + x = 344
x = 344 - 247
x = 97 <=== her score was 97
The answer would be 6. Why? It's because of the 24%. You have to change the 24 into 0.24. If you left it alone, the process would be confusing. Changing it into a decimal is much easier. After you changed the percent into a decimal, multiply the 25. You should get 6 as an answer.
A= -81 + square root of 113
a<span>≈</span> -70.36985418