Standardized Test Prep
SAT/ACT
- Which matrix equation represents the system
{
2
x
−
3
y
=
−
3
−
5
x
+
y
=
14
?
left brace . table with 2 rows and 3 columns , row1 column 1 , 2 x minus , column 2 3 y equals , column 3 negative 3 , row2 column 1 , negative 5 x plus , column 2 y equals , column 3 14 , end table . question mark
-
[
x
y
]
[
2
−
3
−
5
1
]
=
[
−
3
14
]
, matrix with 2 rows and 1 column , row1 column 1 , x , row2 column 1 , y , end matrix . matrix with 2 rows and 2 columns , row1 column 1 , 2 , column 2 negative 3 , row2 column 1 , negative 5 , column 2 1 , end matrix . equals . matrix with 2 rows and 1 column , row1 column 1 , negative 3 , row2 column 1 , 14 , end matrix
-
[
2
−
3
−
5
1
]
[
x
y
]
=
[
−
3
14
]
. matrix with 2 rows and 2 columns , row1 column 1 , 2 , column 2 negative 3 , row2 column 1 , negative 5 , column 2 1 , end matrix , matrix with 2 rows and 1 column , row1 column 1 , x , row2 column 1 , y , end matrix . equals . matrix with 2 rows and 1 column , row1 column 1 , negative 3 , row2 column 1 , 14 , end matrix
-
[
2
−
3
−
5
1
]
[
−
3
14
]
=
[
x
y
]
. matrix with 2 rows and 2 columns , row1 column 1 , 2 , column 2 negative 3 , row2 column 1 , negative 5 , column 2 1 , end matrix . matrix with 2 rows and 1 column , row1 column 1 , negative 3 , row2 column 1 , 14 , end matrix . equals , matrix with 2 rows and 1 column , row1 column 1 , x , row2 column 1 , y , end matrix
-
[
−
3
14
]
[
x
y
]
=
[
2
−
3
−
5
1
]
. matrix with 2 rows and 1 column , row1 column 1 , negative 3 , row2 column 1 , 14 , end matrix , matrix with 1 row and 2 columns , row1 column 1 , x , column 2 y , end matrix . equals . matrix with 2 rows and 2 columns , row1 column 1 , 2 , column 2 negative 3 , row2 column 1 , negative 5 , column 2 1 , end matrix
- What is the value of x if
17
e
4
x
=
85
?
17 , e super 4 x end super , equals 85 question mark
-
5
4
5 fourths
-
In
85
17
⋅
In
4
fraction cap in , 85 , over 17 dot cap in 4 end fraction
-
In
5
4
fraction cap in 5 , over 4 end fraction
-
In
8
5
−
In
17
In
4
fraction cap in 8 5 minus cap in 17 , over cap in 4 end fraction
- A set of data is normally distributed with a mean of 44 and a standard deviation of 3.2. Which statements are NOT true?
- 68% of the values are between 37.6 and 50.4
- 13.5% of the values are less than 40.8
- 5% of the values are lower than 37.6 or higher than 50.4
- I and II only
- I and III only
- II and III only
- I, II, and III
Short Response
-
How can you write the three equations below as a matrix equation for a system? Explain your steps.
2
x
−
3
y
+
z
+
10
=
0
x
+
4
y
=
2
z
+
11
−
2
y
+
3
z
+
7
=
3
x
table with 3 rows and 1 column , row1 column 1 , 2 x minus 3 y plus z plus 10 equals 0 , row2 column 1 , x plus 4 y equals 2 z plus 11 , row3 column 1 , negative 2 y plus 3 z plus 7 equals 3 x , end table
Mixed Review
See Lesson 12-3.
Evaluate the determinant of each matrix.
-
[
−
1
3
7
5
−
4
−
2
0
2
10
]
. matrix with 3 rows and 3 columns , row1 column 1 , negative 1 , column 2 3 , column 3 7 , row2 column 1 , 5 , column 2 negative 4 , column 3 negative 2 , row3 column 1 , 0 , column 2 2 , column 3 10 , end matrix
-
[
17
0
0
0
17
0
0
0
17
]
. matrix with 3 rows and 3 columns , row1 column 1 , 17 , column 2 0 , column 3 0 , row2 column 1 , 0 , column 2 17 , column 3 0 , row3 column 1 , 0 , column 2 0 , column 3 17 , end matrix
-
[
−
3
0
5
5
−
3
2
−
3
−
5
−
2
]
. matrix with 3 rows and 3 columns , row1 column 1 , negative 3 , column 2 0 , column 3 5 , row2 column 1 , 5 , column 2 negative 3 , column 3 2 , row3 column 1 , negative 3 , column 2 negative 5 , column 3 negative 2 , end matrix
See Lesson 11-6.
Find the mean, variance, and standard deviation for each data set.
- 29, 35, 44, 25, 36, 30, 40, 33, 38
- 5.2, 6.0, 3.5, 4.4, 2.5, 3.0, 4.6
- 14 m, 18 m, 22 m, 28 m, 15 m, 21 m
- 71 mi, 60 mi, 82 mi, 30 mi, 44 mi
Get Ready! To prepare for Lesson 12-5, do Exercises 70–72.
See Lesson 2-7.
Graph each equation. Then describe the transformation from the parent function
f
(
x
)
=
|
x
|
.
f open x close equals vertical line x vertical line .
-
f
(
x
)
=
|
x
+
4
|
f open x close equals vertical line x plus 4 vertical line
-
f
(
x
)
=
|
x
|
−
3
f open x close equals vertical line x vertical line negative 3
-
f
(
x
)
=
|
x
−
5
|
+
3
f open x close equals vertical line x minus 5 vertical line plus 3
