Answer:
A. The slope of Function A is greater than the slope of Function B.
Step-by-step explanation:
The slope of a function can be defined as rise/run. In Function A, the rise/run is 4. The slope in Function B is much easier to see: it is 2. Because 4 is greater than 2, Function A has a greater slope than Function B.
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).
Answer:
(a) The probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.
(b) <em>n</em> = 28.09
Step-by-step explanation:
The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.
It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.
The mean and standard deviation of <em>Y</em>ₙ are:
(a)
For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:
Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:
**Use a <em>z </em>table for probability.
Thus, the probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.
(b)
It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.
The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.
Compute the value of <em>n</em> as follows:
The last equation is a quadratic equation.
The roots of a quadratic equation are:
a = 16
b = -805.4289
c = 10000
On solving the last equation the value of <em>n</em> = 28.09.
We have to check which of PEMDAS rules can be applied in this case. We have only ADDITION. In this case we can observe that addition is:
1. Associative
2. Commutative
3. has Additive property
