D because n= -14 and were dividing 56
The answer is. $1500.
Let x represent the amount invested at 5%, then is the amount invested at 8%, and is the amount invested at 6%. Since the total return is $1270:
let money invested at 5%= x
money invested at 6%
= 20000- (X+X-1500)
21500-2x
Total interest for 1 year.
x×5/100+ (x-1500)×8/100+(21500-2x)×6/100 = 1270
5x/100+8(x-1500)/100+(21500-2x)6/100=1270
5x+8x-12000+129000-12x = 127000
x+117000 = 127000
x = 10000
money invested at 8%
=21500-2×10000
= 1500.
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Average (mean) = (sum of all the data) / (# of data)
sum of all the data = (average)(# of data)
Thus for 100 students with an average of 93,
sum of all data = (93)(100) = 9300
and for 300 students with an average of 75,
sum of all data = (75)(300) = 22500
Therefore you would expect the overall average to be
(9300 + 22500) / (100 + 300) = 79.5 %
Now if there are x # of advanced students and y # of regular students, then
x + y = 90 (total # of students) and 93x + 75y = 87(x + y) (overall average)
The second equation can be simplified to x - 2y = 0
Subtracting the two equations yields
x = 60 and y = 90
Therefore you would need 60 advanced and 30 regular students.
The answer to this question is c=3
Solution set of inequality-7<3x+2<11, where x is a real number
