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 ) =
12
3
=.
52
13
(b) There are 13 diamonds plus three 7s
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' F A rDU4dBii
7
cfw_ | cfw_%
s3 z cfw_
biE tbE F A z
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 rank. In this case, that is the 34th ranked number; thu
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 of 6 mice recover) =
6
(0.70)3 (0.30)3 .185.
3
83
ways
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
security numbers. Since there are only 281 million americans,
G p t S G@G @X t B I WX @G h R I iSDXD G dD Gr
a cp `qbfC&cHEDee!c&1ecHTH%eHP6IB | E&ceCtfC&H866T88
iG BDrG ftG@ a IS a a
h S D IS iSG@D IX a XD IX@X R X W G@G @X I@S WX tD I X RS G dD GDX IG
SHDCk6TaTfCHcCcHc`CY@&HPcAHT`TxecQT6CQCceP
7
6`EXHcCQ6c6nED
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 $3 per guide sold. Therefore, R(x) = 3x. In order to b
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
x y = 3
5y + z = 7
3y 2z = 12
3R2 + 5R3 R3
to nd
x y =
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 question. Thus 4 = (1) 2 + b This gives
b = 6. The equation of
1S $
DsU0'G
(W G"W G G " ( 1W % ) $ X ) f W ( X ) ) ( ( " X )S %" 8 4
gsg0g'q'ggp&)HT`'&)d(&$T2pH0s`!w'&)pYV`w1'V`!0w!&653
8 y u y y 8 y Au yuu X ) ) 9 1 G CA
EYs0dv&rBup0RTsv0dfQlvQ@nlIe'skEjS9
y y z A e ) 11 9 z A u y e ) W" 9 1 G CA
sb~0ere#0d2ggsTl
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 ) P (Q|R3 )
(.40)(.05)
=1
(.40)(.05) + (.30)(.60) + (.6
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). Thus there are 30 dierent ways to
4
select this committ
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 0.1539. So the
percent between the z-scores is 0.1539 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 1 , D = .6 .1 .3 , D = .36 .15 .49
.6 .1 .3
.36 .15 .
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 eliminate the x-terms from R1 and R2 .
R1
2R1 R2 R2
3R1 R3 R3
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 system.
R1
R2
121
243
Use row operations to put the system
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 we plug in the point (4, 3) to nd b:
y = mx + b
3 = (1)
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 (R2 |Q) =
2
7.6.10 Let Q be the event the person is qual
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!
2
8.2.14 Find the number of ways of choosing 2 letters
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. Find the z -score such that 25% of the total area is to
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
49-58
59-68
69-78
79-88
89-98
6
13
20
19
13
9
20
15
10
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 experiment
is repeated twice, and on each trial there a
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 card is a
face card. Then n(F ) = 12 since there are 12
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 is 1.
10.1.18
The transition matrix associated to this
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]. The fundamental matrix is
9
then given by F = (I1 Q)1 = (
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 variable x is the number of cases of EZ-CHEEZ produced, an
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 b has equation y = mx + b. A line with slope m passing