MATH 105: Assignment 10 Solutions
Each problem worth 4 points
9.3.10. The area between z = .64 and z = .211 is
P (.64 < z < 2.11) = P (z < 2.11) P (z < .64)
= .9826 .7389
= .2437 or 24.37%.
2
9.3.18.
MATH 105: Assignment 4 Solutions
Each problem worth 4 points
7.4.16
(a) The cards less than 4 include 2s, 3s, and aces. There are a total of 12
of these cards in a deck of 52, so
P ( ace or 2 or 3 ) =
Math 105 Prelim #3 Solutions
1.
(a) The mode is the most frequently occuring data point, which is 38. (It occurs 18 times.) (b) The median is the middle number when the data are ordered in increasing
Math 105
Fall 2003
Solutions
Assignment 8 [100pts]
6
ways to select the three mice that will recover. Each way
3
occurs with probability (0.70)3 (0.30)3 . Thus,
8.4 #50 [10 pts] There are
P (exactly 3
MATH 105 - HOMEWORK 7
Section 8.1
46) There are 10 choices for the rst spot, 10 for the second, and so on. By
the multiplication principle, there are therefore 109 = one billion possible social
securi
Math 105 Final Exam (Fall 2003) Solutions
1.
(a) There is a xed cost of $50, and a variable cost of $2 per guide. Let x be the number
of guides sold. Therefore,
C(x) = 50 + 2x.
(b) Spencers revenue is
Math 105
Fall 2003
Solutions
Assignment 3 [100pts]
Review page 117-122 #6 [6 pts] Begin with the system
x y = 3
2x + 3y + z = 13
3x 2z = 21
and perform
2R1 + R2 R2 and 3R1 + R3 R3
to yield
followed by
Math 105
Fall 2003
Solutions
Assignment 1 [100pts]
1.1 16 [10pts] We seek the equation of the line in the form y = mx + b. We know that m = 1
(the slope) and the point (2, 4) is on the line in questio
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Math 105
Fall 2003
Homework 5 Solutions
[100 points]
Section 7.6
#6 [10 pts]. We have P (R1 |Q) = 1 P (R1 |Q), and so by Bayes Theorem,
P (R1 ) P (Q|R1 )
P (R1 ) P (Q|R1 ) + P (R2 ) P (Q|R2 ) + P (R3
Math 105
Fall 2003
Solutions
Assignment 6[100pts]
8.2 12a[5pts] Selecting a committee of 4 among 30 members is choosing a subset of 4 out of
the set of 30 (no repetitions, the order does not matter).
k S E# G ) 3 ( s 2 2 3 2 s
hbb1hvF1qsBii"q4Bqhde1"m
!E A !uE !h!E A uh!h
% S ES E% E
7
w # 4# G ) 3 E 4 h# ( 4 ( 4 f 3 2# Y W4
'rVb1Fe11XdqiXd0v(cXt#!iivi~8gw1T Q @986
k k z y x z y 2 3
FdSk A z d
Math 105
Fall 2003
Homework 10 Solutions
[100 points]
Section 9.3
#12 [10 pts]. According to the normal curve table, the area to the left of
z = 1.74 is 0.0409, and the are to the left o z = 1.02 is 0
Math 105
Fall 2003
Solutions
Assignment 11[100pts]
10.1 14[5pts] Yes, it is a transition matrix.
1
n
b
1
3
1
3
'
a
n
T
1
2
c
E
1
3
1
n
T 2
10.1 22[10pts]
.3 .2 .5
.39 .11 .5
.417 .128 .455
2
3
D = 0 0
MATH 105: Assignment 3 Solutions
Each question worth 4 points
Ch. 2 Review #6
We wish to solve the system of three linear equations:
x y + 0z = 3
2x + 3y + z = 13
3x + 0y 2z = 21
R1
R2
R3
First elimin
MATH 105: Assignment 2 Solutions
Each problem worth 4 points
2.2.22
We wish to solve the following system of two linear equations:
x + 2y = 1
2x + 4y = 3.
Write down the augmented matrix for this syst
MATH 105: Homework 1 Solutions
Each problem worth 4 points
1.1.20
Find the equation y = mx + b of the line through (8, 1) and (4, 3).
The slope is given by rise over run:
m=
3 (1)
4
=
= 1.
48
4
Then w
MATH 105: Assignment 5 Solutions
Each problem worth 4 points
7.6.4
P (R2 ) P (Q|R2 )
P (R1 ) P (Q|R1 ) + P (R2 ) P (Q|R2 ) + P (R3 ) P (Q|R3 )
.6(.3)
=
.05(.4) + .6(.3) + .35(.6)
18
.18
=
=
.41
41
P (
MATH 105: Assignment 6 Solutions
Each problem worth 4 points
8.2.10 We wish to choose a hand of 6 clubs, from a total of 13 clubs. The
number of dierent ways we can do this is
13
6
=
13!
= 1716.
6!7!
MATH 105: Assignment 9 Solutions
Each problem worth 4 points
9.1.4. Here is the frequency distribution and histogram. Youll have to imagine the frequency polygon in your mind.
Interval Frequency
39-48
MATH 105: Assignment 8 Solutions
Each problem worth 4 points
8.4.22. The properties of a binomial experiment are listed on page 424.
Clearly the rst two properties are satised by this experiment (the
MATH 105: Assignment 7 Solutions
Each problem worth 4 points
Chapter 7 Review Exercise 44. Let S be the sample space of all 52 cards,
let J be the event the card is a Jack, and let F be the event the
MATH 105: Assignment 11 Solutions
Each problem worth 4 points
10.1.12
Yes,
1
4
1
2
3
4
1
2
is a transition matrix since
It is a square matrix
Entries are between 0 and 1.
Sum of entries in each row
SECTION 10.3
8) [10 pts] First, we rearrange the transition matrix so that the absorbing
states (states 1 and 3) come rst:
1 0 0
1
3 0 1 0
2
.6 .3 .1
So now, we have that R = [.6 .3] and Q = [.1]. Th
Math 105 Prelim #1 Solutions
1.
(a) ii. m is negative, but we do not know the exact value of m (b) iii. most of the data points are close to the least squares line
2.
(a) Note that the independent var
Math 105 Final Exam Review
The nal exam will take place on Thursday, December 11, 2003, from 9:00
11:30 am in Warren Hall B45.
Slopes and equations of lines (1.1). A line with slope m and y-intercept