A value that "lies outside" (is much smaller or larger than) most of the other values in a set of data.
For example in the scores 25,29,3,32,85,33,27,28 both 3 and 85 are "outliers".
Y = kx
Plug in what we know:
5 = k(8)
5 = 8k
Divide 8 to both sides:
k = 0.625
Plug this back into the equation along with y = 15:
y = kx
15 = 0.625x
Divide 0.625 to both sides:
x = 24
155 divided by 17 =
9.1176
Answer:
(3x+4)(5x+7)
Step-by-step explanation:
15x^2
+41x+28
Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2
+ax+bx+28. To find a and b, set up a system to be solved.
a+b=41
ab=15×28=420
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.
1,420
2,210
3,140
4,105
5,84
6,70
7,60
10,42
12,35
14,30
15,28
20,21
Calculate the sum for each pair.
1+420=421
2+210=212
3+140=143
4+105=109
5+84=89
6+70=76
7+60=67
10+42=52
12+35=47
14+30=44
15+28=43
20+21=41
The solution is the pair that gives sum 41.
a=20
b=21
Rewrite 15x^2
+41x+28 as (15x^2
+20x)+(21x+28).
(15x^2
+20x)+(21x+28)
Factor out 5x in the first and 7 in the second group.
5x(3x+4)+7(3x+4)
Factor out common term 3x+4 by using distributive property.
(3x+4)(5x+7)
