Answers to the Assignment Questions:
1. e2x – 9ex + 14 = 0
If z = ex
This can be written as:
z2 – 9z + 14 = 0
Get the values of z:
(z -2) (z-7) = 0
z = 2 and z = 7
Bringing back in terms of ex yields,
ex = 2 and ex = 7
Get the natural log of both sides will be:
ln(ex = 2) and ln ex = 7
x = ln 2 and x = ln 7
2. Solve the equation:
log (x-1) + log (x+1) = 2
log (x – 1) (x + 1) = 2
log (x2 -1) = 2
log10 (x2 – 1) = 2
102 = x2 -1
x2 – 101 = 0
Using the quadratic formula the value of x can be computed:
- b+ √b2 – 4ac / 2a
√-4(-101) / 2(1)
√404 / 2
x = √101 The negative (-√101)is not the solution since, there is no negative value for log.
To check, substitute the value to the original equation.
log (√101-1) + log (√101+1) = 2
2 = 2 True
3. Find the AB and BA when
1 2 1 2 4 1
A = 3 4 0 B = 0 3 1
0 1 -1 0 0 4
This can yield to:
For the BA:
This yield to:
4. Solving the equations:
x + 2y = 15
x – 2y = – 5
The first step is to convert into matrix.
[3 4 : 5]
[2 -1 : 7]
Change into identity matrix by getting zero in the Row 1, Column 2. So add (1 x row 2) to row 1.
[2 0 : 10] added (1 x row 2) to row 1
[1 -2 : -5]
Next is want 1 in the top left corner.
[1 0 : 5 ] Divided row 1 by 2
[1 -2 : -5]
Next is to have zero in the bottom left corner.
[1 0 : 5]
[0 -2 : -10] added (-1 x row 1) to row 2
Finally, is zero to bottom left corner.
[1 0 : 5]
[0 1 : 5] divided by -2
Then, it has the 2 x 2 matrix which can read as:
x = 5
y = 5
5. The data can be tabulated as:
Grades of the Students
Frequency
0 – 0.5
2
0.6 – 1
2
1.1 – 1.5
2
1.6 – 2
9
2.1 – 2.5
7
2.6 – 3
5
3.1 – 3.5
7
3.6 – 4
6
The frequency histogram is given below:
The frequency Polygon is given below:
6. The O,give for the data is given below:
7.
The stem and leaf display is given below:
9
14
8
006
7
0235
6
468
a. The mean for the data is 66 + 70 + 68 + 75 + 80 + 86 + 64 + 73 + 80 + 94 + 72 + 91
12
The mean = 919 / 12 = 76.58
b. The median is the middle of 64, 66, 68, 70, 72, 73, 75, 80, 80, 86, 91, and 94.
The median is 73 + 75 / 2
The median = 74
c. Mode occurrence of data most often which is 80.
8. The expression can be written as:
(x1 y1) + (x2 y2) + (x3 y3) + (x4 y4) +(x5 y5)
9.
a. The range of the grades is 94 – 64 = 30
b. The standard deviation can be computed as:
Frequency
x – mean(76.58)
{x - mean(76.58)}squared
66
10.58
111.9364
70
6.58
43.2964
68
8.58
73.6164
75
1.58
2.4964
80
-3.42
11.6964
86
-9.42
88.7364
64
12.58
158.2564
73
3.58
12.8164
80
-3.42
11.6964
94
-17.42
303.4564
72
4.58
20.9764
91
-14.42
207.9364
Sum
1046.917
Using the formula:
s = √ ∑(x – xbar)2 / n -1
s = √1046.917 / 11
s = √95.17
s = 9.76
c. The variance has the formula of:
s2 = ∑(x – xbar)2 / n -1
s2 = 1046.917 / 11
s2 = 95.17
Credit:ivythesis.typepad.com
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