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.
_____
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.)
AVERAGE between 90 and 100%
means not including 90 and 100% but not including 90 or 100%
average=(sum of values in set)/(how many values in the set)
so
4 exams, so there are 4 values
average=(sum)/4
the sum is the known+unknwon
sum=76+99+86+x
so
90<span><</span>average and average <span><</span>100
or
90<span><</span>average<span><</span>100
lets say average=a
a=sum/number
a=(76+99+86+x)/4
90<(76+99+86+x)/4<100
90<(76+99+86+x)/4 and (76+99+86+x)/4<100
solve each for x and find intersection
90<span><</span>(76+99+86+x)/4
times 4 both sides
360<span><</span>261+x
minus 261 both sides
99<span><</span>x
(76+99+86+x)/4<u><</u>100
times 4 both sides
261+x<u><</u>400
minus 261 both sides
x<u><</u>139
so
99<u><</u>x<u><</u>139
since max score is 100
99<u><</u>x<u><</u>100
interval notaion is
[99,100]
answer is 2nd one
I think the answer would be the first one
Answer:
8. 14 for the third question.
9. 18 for the forth question.
Step-by-step explanation:
the third and forth orange equations.
