Kimberly Ragnauth
BioStats HW#3
February 17th, 2016
Section A08
Chapter 8
2.) a. Negative
b.This positive association can be accounted for the fact that the owners have higher
income. Since they have higher incomes they can afford cars that are more fuel
Kimberly Ragnauth
BioStats 148 HW #1
January 31st, 2016
Section A08
Chapters 1&2
3.) The parents who refused to allow their children to participate in the experiment were
not correct in thinking that their child would have a higher risk of polio. This is
Kimberly Ragnauth
BioStats HW#2
February 7th, 2015
Section A08
Chapter 4
5.) 80mm is unusually low.
115mm is average.
120mm is average.
210 is unusually high.
6.) a. 40- iii
50- ii
60- i
b. median is less than average- iii
median is about equal to the ave
Kimberly Ragnauth
BioStats HW#6
March 6th, 2016
Section A08
Chapter 14
1. a. P(both show 3)= () x ()= 1/36,
b. P(same number)= (1/36) x (1/36) x (1/36) x (1/36) x (1/36) x (1/36)= 6/36
4. Option one is better because by having the chance of having at leas
Kimberly Ragnauth
BioStats HW#4
February 24th, 2016
Section A08
Chapter 10
1.) (i)- A
(ii)- C
(iii)- B
(iv)- Not given.
2.) a.) The estimated average score at age 35 for all the individuals who scored 115 at age 18 is
112.
b.) The predicted score is 112 a
Kimberly Ragnauth
BioStats HW#11
April 21st, 2016
Section A08
Chapter 22
2.)
Chapter 23
1.) The expected value for the draws will be 100. The SD is 20. The SE for the sum of draws will
be x 20= 400. The SE for the average is 400/400=1. You need to then co
Kimberly Ragnauth
Biostats HW#10
April 13th, 2016
Chapter 20
4.) There is a 69% chance that the sum of the gross income is over $33,000,000.
7.) A. True
B. False
C. True
D. False
E. False
F. False
9.) 200
11 or 12
12.) 952
37 or 38
Chapter 21
2.) a. Fract
Kimberly Ragnauth
BioStats HW#5
February 28th, 2016
Section A08
Chapter 12
4.) a. 1
b. The line is not the regression line because based on the data the regression line would
slope downwards. This line could probably be the average value of the Y variable
4.3. # 2.
Proof.
(a). We first prove one implication.
Suppose that X(0) = 0. Then X(x) = aX + b. Thus
X (x) = a, 0 x l.
Hence the boundary conditions,
X (0) a0X(0) = 0,
X (l) + alX(l) = 0.
Thus we have
a a0b = 0,
a + al(al + b) = 0.
Hence
(a0 + al)b = a0a
P45.
2.3 #3.
Proof.
(a). By the strong maximum principle, u(x, t) > 0 in the interior points 0 < x < 1, 0 < t < because the
minimum value of u, 0, is attained at the boundary point, and at the two lateral sides.
(b). We follow the hint. Let (t) = the maxi
P45.
2.3 # 2.
Proof.
(a). Let M(T) = the maximum of u(x, t) in the closed rectangle cfw_0 x l, 0 t T. M(T) is a decreasing
function of T. Indeed, when T1 T2, the rectangle R1 = cfw_0 x l, 0 t T1 is contained in the rectangle
R2 = cfw_0 x l, 0 t T2; the bo
P45.
2.3 # 1.
Proof.
By the maximum principle, u(x, t) = 1 x 2 2kt attains the maximum at the bottom or on the two sides.
When t = 0, 1 x 2 2kt = 1 x 2 attains the maximum at x = 0, i.e., 1 x 2 = 1. When x = 0, 1 x 2 2kt
= 1 2kt attains the maximum 1 in t
P41.
2.2 # 4.
Proof.
The equation utt = uxx has a solution
u(t, x) = f(x + t) + g(x t), 6 where f and g are functions of one variable.
Thus
u(x + h, t + k) = f(x + t + h + k) + g(x t + h k), u(x h, t k)
= f(x + t h k) + g(x t h + k), u(x + k, t + h)
= f(x
P41.
2.2 # 3.
Proof.
(a). The solution u satisfies utt = uxx. For the translate of u = u(x y, t), we differentiate both sides in t
and x,
utt(x y, t) = uxx(x y, t).
(b). Differentiate both sides of utt = uxx in x in any order,
k
k
( x u)tt = ( x u )xx.
Di
P38.
2.1# 11.
Proof. For this equation, we have a particular solution
u(x, t) =
1
16
sin(x + t).
Then we look for solutions to the homogeneous equation
(3x + t)(x + 3t)v = 0.
Then in Lemma 6.2,
a = 3, b = 1, c = 1, d = 3.
Then by (6),
v(x, t) = f(x 3t) +
P24.
1.5 # 2.
Proof.
(a). Fouriers law says that the heat flows from hot to cold regions proportionately to the temperature
gradient. So if the metal rod insulated at the end x = 0, the temperature at x = 0 is
u
x
= 0.
(b). Ficks law says that a chemical
Exponential Functions and Dierentiability
R. Michael Range
State University of New York at Albany, Albany, New York 12222
E-mail address : [email protected]
Contents
1. Exponential Functions
1.1. Compound interest
1.2. The Functional Equation
1.3. Den
Math 424-524
Exam 1
Fall 2012
1. Let A Mn (F ) and v F n . Let k be the smallest positive integer
such that v, Av, A2 v, . . . , Ak v are linarly dependent.
(a) Show that we can nd a0 , . . . , ak1 F with
a0 v + a1 Av + + ak1 Ak1 v + Ak v = 0
(note that t
Math 424-524
Exam 2
Fall 2012
Let F be a eld. We say that A Mn (F) is periodic , or of nite order ,
if Ak = In for some k > 0. Note this implies that A is invertible with
inverse Ak1 . So A GLn (F), the group of n n invertible matrices with
coecients in F
Math 362
Exam 1 solutions
Fall 09
1. Urn A contains 7 red balls and 3 green balls. Urn B contains 4 red
balls and 6 green balls. An experiment is conducted as follows:
Roll a pair of dice. If the sum is 5 draw three balls from
Urn A without replacement. O