Answer:
35/79
Step-by-step explanation:
Assuming you can't choose the same song a second time, the probability is 35/79.
If you can choose the same song, the probability is 36/80, which reduces to 9/20.
Answer:
- vertical scaling by a factor of 1/3 (compression)
- reflection over the y-axis
- horizontal scaling by a factor of 3 (expansion)
- translation left 1 unit
- translation up 3 units
Step-by-step explanation:
These are the transformations of interest:
g(x) = k·f(x) . . . . . vertical scaling (expansion) by a factor of k
g(x) = f(x) +k . . . . vertical translation by k units (upward)
g(x) = f(x/k) . . . . . horizontal expansion by a factor of k. When k < 0, the function is also reflected over the y-axis
g(x) = f(x-k) . . . . . horizontal translation to the right by k units
__
Here, we have ...
g(x) = 1/3f(-1/3(x+1)) +3
The vertical and horizontal transformations can be applied in either order, since neither affects the other. If we work left-to-right through the expression for g(x), we can see these transformations have been applied:
- vertical scaling by a factor of 1/3 (compression) . . . 1/3f(x)
- reflection over the y-axis . . . 1/3f(-x)
- horizontal scaling by a factor of 3 (expansion) . . . 1/3f(-1/3x)
- translation left 1 unit . . . 1/3f(-1/3(x+1))
- translation up 3 units . . . 1/3f(-1/3(x+1)) +3
_____
<em>Additional comment</em>
The "working" is a matter of matching the form of g(x) to the forms of the different transformations. It is a pattern-matching problem.
The horizontal transformations could also be described as ...
- translation right 1/3 unit . . . f(x -1/3)
- reflection over y and expansion by a factor of 3 . . . f(-1/3x -1/3)
The initial translation in this scenario would be reflected to a translation left 1/3 unit, then the horizontal expansion would turn that into a translation left 1 unit, as described above. Order matters.
Answer:
44%
Step-by-step explanation:
1) You divide 209 by 475
2) Take 0.44 and move the decimal over 1 place to make <u>44%</u>
Answer:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z=1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).
-18 hope this doesn’t get hate from people that a gods at this app and don’t know how to mind there beeswax