Download Top of Form Write the first five terms of the arithmetic sequence: a1

Write the first five terms of the arithmetic sequence: a1 = 3; d = 3
3, 5, 7, 9, 11
6, 9, 12, 15, 18
0, 3, 6, 9, 12
3, 6, 9, 12, 15
For an arithmetic sequence, an 1  an  d
a2  a1  d  3  3  6,
a3  a2  d  6  3  9
a4  a3  d  9  3  12, a5  a4  d  12  3  15
A spinner is used for which it is equally probable that the pointer will land on any one of
six regions. Three of the regions are colored red, two are colored green, and one is
colored yellow. If the pointer is spun twice, find the probability it will land on green and
then yellow.
2 1
The probability of landing on green is: 
6 3
The probability of landing on yellow is:
1
6
So the probability of landing on green then yellow is
1 1 1
 
3 6 18
1/18
1/3
1/6
1/9
3 of 25
Use properties of limits to find the indicated limit. It may be necessary to rewrite an
expression before limit properties can be applied:
2x-7
limx  1
4x+5
2 x  7 2 *1  7
5


x 1 4 x  5
4 *1  5
9
lim
-5/9
–1/2
does not exist
–7/5
4 of 25
A piecewise function is given. Use the function to find the indicated limits, or state that a
limit does not exist.
(a) lim is over xd – f(x), (b) lim is over xd + f(x), and (c) lim is over xd f(x)
f(x) =
; d = -3
2
When x  0 , f ( x)  x  5
When x  3  or -3, or 3 , x 2  9
Therefore, f ( x)  9  5  4
(a) -5 (b) -2 (c) does not exist
(a) 4 (b) 4 (c) 4
(a) -2 (b) -5 (c) does not exist
(a) -5 (b) -2 (c) -2
5 of 25
A restaurant offers a choice of 5 salads, 9 main courses, and 3 desserts. How many
possible 3-course meals are there?
There are C51  5 ways of choosing salad, C91  9 ways of choosing main courses, and
C31  3 ways of selecting desserts. So the possible 3-course meals are:
C51C91C31  5 * 9 * 3  135
17 possible meals
135 possible meals
45 possible meals
270 possible meals
6 of 25
Evaluate the given binomial coefficient:
12
( )
3
12!
12  11  10  9  8  7  6  5  4  3  2  1 12  11  10



 220
3!(12  3)!  3  2  1 9  8  7  6  5  4  3  2  1
3  2 1
79,833,600
4
220
110
7 of 25
Use the graph of f to find the indicated limit
limx  f(x)
It can be observed that f ( x)  0 when x  
3
4
does not exist
0
8 of 25
Evaluate the given binomial coefficient:
11
( )
7

11!
11  10  9  8  7  6  5  4  3  2  1 11  10  9  8  7  6  5


 330
7!(11  7)!  7  6  5  4  3  2  1 4  3  2  1
7  6  5  4  3  2 1
7920
1
165
330
9 of 25
How many 2-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0?
No digit can be used more than once.
For the first digit, there are 9 choices since 0 cannot be in the first place. Then there are
10 – 1 = 9 choices since no digit can be used more than once. So the 2-digit numbers:
C91C91  9 * 9  81 . There is no correct solution in the selections.
From 10 to 99, there are total 90 numbers. Then we need to exclude 11, 22, 33, 44, 55,
66, 77, 88, 99, which are total of 9 numbers. So the total is 90 – 9 = 81.
90
45
1,814,400
3,628,800
10 of 25
Use properties of limits to find the indicated limit. It may be necessary to rewrite an
expression before limit properties can be applied.
x4-1
limx  1
x-1
x  1  0 when x  0 . So we cannot substitute x to the function to evaluate the limit.
x 4  1 ( x 2  1)( x 2  1) ( x  1)( x  1)( x 2  1)
when x  1,


 ( x  1)( x 2  1)
x 1
x 1
x 1
x  1 implies that x is not to equal 1, therefore,
x4  1
 lim  x  1 x 2  1  (1  1)(1  1)  4
x 1 x  1
x 1

lim
0
does not exist
4

2
11 of 25
Use the empirical probability formula to solve the exercise. Express the answer as a
fraction. Then express the probability as a decimal, rounded to the nearest thousandth, if
necessary.
The table below represents a random sample of the number of deaths per 100 cases for a
certain illness over time. If a person infected with this illness is randomly selected from
all infected people, find the probability that the person lives 3-4 years after diagnosis.
Years after Diagnosis
Number of Deaths
1-2
15
3–4
35
5–6
16
7–8
9
9 – 10
6
11 – 12
4
13 – 14
2
15+
13
Out of total 100 deaths, there are 35 people lives 3-4 years after the diagnosis, so the
35
7

 0.35
probability is
100 20
7/120; 0.058
35/100; 0.35
1/35; 0.029
35/65; 0.538
12 of 25
If the given sequence is a geometric sequence, find the common ratio:
4, 12, 36, 108, 324
a
a 12
r  n 1  2 
3
an
a1
4
not a geometric sequence
12
3
1/3
13 of 25
Evaluate the factorial expresion:
9! 9  8  7  6  5  4  3  2  1

 9 *8  72
7!
7  6  5  4  3  2 1
9!/7!
9/7
72
2!
9
14 of 25
You are dealt one card from a standard 52-card deck. Find the probability that you are not
dealt a 10.
4
1

There are four 10 in the card deck. The probability of dealing a 10 is:
52 13
So the probability of not dealing a 10 is 1 
1/13
9/10
1/10
12/13
15 of 25
1 12

13 13
Find the slope of the tangent line to the graph of f at the given point:
f(x) = -2x + 6 at ( 2, 2)
The slope of the tangent line is m  f '( x)  2
Actually, the tangent line of a straight line is the line itself.
-6
6
-2
-1/2
16 of 25
Find the derivative of f at x. That is, find f'(x).
f(x) = x3 + 19; x = 8
d 3
f '( x) 
x  19  3x31  0  3x 2
dx
f '(8)  3*82  192


24
1536
192
6
17 of 25
The function f(x) = x3describes the volume of a cube, f(x), in cubic inches, whose length,
width, and height each measure x inches. If x is changing, find the average rate of change
of the volume with respect to x as x changes from 4 inches to 4.1 inches.
Average rate of change
=
f
f ( x  x)  f ( x) f (4.1)  f (4) 4.13  43



 49.21 cubic inches per inch
x
x
4.1  4
0.1
If x is small enough, so x  0 , the rate of change is f '( x)  lim
x  0
f ( x  x)  f ( x)
x
33.23 cubic inches per inch
1329.21 cubic inches per inch
-49.21 cubic inches per inch
49.21 cubic inches per inch
18 of 25
Evaluate the given binomial coefficient:
9
( )
6
9!
9  8  7  6  5  4  3  2 1
9 *8* 7
=


 84
6!(9  6)!  6  5  4  3  2 13  2 1 3  2 1
42
504
84
1
19 of 25
Use the formula for the value of an annuity to solve the problem. Round your answer to
the nearest dollar.
Looking ahead to retirement, you sign up for automatic savings in a fixed-income 401K
plan that pays 6% per year compounded annually. You plan to invest $3500 at the end of
each year for the next 20 years. How much will your account have in it at the end of 20
years?
The future value of an annuity: FVoa = PMT [((1 + i)n - 1) / i]
Here, PMT = 3500, interest I = 6% = 0.06, n = 20
So FV  3500
(1  0.06) 20  1
 128749.57  128750
0.06
$130,520
$127,207
$128,750
$130,048
20 of 25
Write a formula for the general term (the nth term) of the arithmetic sequence. Do not use
a recursion formula. Then use the formula for an to find the indicated term of the
sequence.
Find a11; 12, 10, 8, ...
an  an 1  d
an 1  an 1  d
a2  a1  d
Add all equations together, an  a1  (n  1)d
In the sequence, 12, 10, 8, a1 = 12, d = a2 – a1 = 10 – 12 = 2
a11  a1  (11  1)d  12  10 * (2)  8
.32
-8
.-10
.34
21 of 25
Complete the table for the function and find the indicated limit:
limx2 x2 + 8x - 12
x
1.9
1.99
f(x)
6.81
7.880
1.999 2.001
2.01
2.1
7.988 8.012 8.120 9.210
When x  2, f ( x)  8.0
6.810; 7.880; 7.988; 8.0120; 8.120; 9.210; limit = 8.0
16.810; 17.880; 17.988; 18.012; 18.120; 19.210; limit = 18.0
16.692; 17.592; 17.689; 17.710; 17.808; 18.789; limit = 17.70
5.043; 5.364; 5.396; 5.404; 5.436; 5.763; limit = 5.40
22 of 25
Write the first five terms of the sequence whose general term is given:
an = 4(2n - 3)
a1  4(2 *1  3)  4, a2  4(2 * 2 *3)  4, a3  4(2 *3  3)  12
a4  4(2 * 4  3)  20, a5  4(2 *5  3)  28
-1, 1, 3, 5, 7
-4, -8, -12, -16, -20
-12, -4, 4, 12, 20
-4, 4, 12, 20, 28
23 of 25
Use the formula for the general term (the nth term) of a geometric sequence to find the
indicated term of the sequence with the given first term, a1, and common ratio, r.
Find a12 when a1 = 2, r = 2
an  a1r
n 1
a12  2 * 212 1  212  4096
4100
4096
24
8192
24 of 25
Use the Binomial Theorem to expand the expression and express the result in simplified
form.
(3x + 2)3
n
n
    a n kbk
k 0  k 
3
 3
 3
 3
 3
 3
(3 x  2)3     (3 x)3 k (2) k    (3 x)3 0 (2) 0    (3 x)3 1 (2)1    (3 x)3  2 (2) 2    (3 x) 3  3 (2) 3
k 0  k 
0
1
 2
 3
 1(3 x)31  3(3 x) 2 2  3(3 x)(2) 2  1*1(23 )
a  b
n
 27 x3  54 x 2  36 x  8
27x3 + 54x2 + 36x + 8
9x2 + 12x + 4
27x3 + 54x2 + 54x + 8
9x6 + 6x3 + 64
25 of 25
An explosion causes debris to rise vertically with an initial velocity of 9 feet per second.
The functions (t) = -16t2 + 144t describes the height of the debris above the ground, s(t),
in feet, t seconds after the explosion. What is the instantaneous velocity of the debris 5
second(s) after the explosion?
dh d
v(t ) 

16t 2  144t  16 * 2t  144  32t  144
dt dt
v(5)  32 * 5  144  16

-160
-16
16

160