Honors Discrete Chapter 10 Test Review Guide

Honors Discrete Chapter 10 Test Review Guide - SOLUTIONS

SECTION 10.2 – LINEAR GROWTH MODELS Practice Problems

For each of the following linear growth model,

1)  P0 = 718 and d = 32

PN = 718 + 32N;

P9 = 1006

2)  3, 19, 35, 51, …

PN = 3 + 16N;

P9 = 147

3)  P5 = 97 and P6 = 86

PN = 152 - 11N; d = 86 – 97 = -11

P9 = 53

4)  P13 = 217 and P14 = 228

PN = 74 + 11N; d = 228 – 217 = 11

P9 = 173

5)  P6 = 165 and P15 = 318

PN = 63 + 17N;

P9 = 216

6)  P21 = 1896 and P90 =4449

PN = 1119 + 37N;

P9 = 1452

For each of the following linear growth model, find the given sum

7)  8 + 14 + 20 + 26 + 32 + 38 + 44 + 50

232 =

8)  For 10 terms, 11 + 27 + 43 + 59 + …

830 =

P9 = 11 + 16(9) = 155

9)  P0 = 115 and d = 83;

9a. P0 + P1 + P2 + … + P29

39,555 =

P29 = 115 + 83(29)

9b. P21 + P22 + … + P30 + P31

25,003 =

P21 = 1858 = 115 + 83(21)

P31 = 2688 = 115 + 83(31)

10) For 50 terms, 37 + 44 + 51+ …

10,425 =

P49 = 37 + 7(49) = 380

11) 17 + 29 + 41 + … + 1157

56,352 =

1157 = 17 + 12(95) à P95 = 1157

0 à 95 = 96 #’s

12)  8895 + 8778 + … + 7140

128,280 =

7140 = 8895 - 117(15) à P15 = 7140

0 à 15 = 16 #’s

13) P3 = 127 and P16 = 400;

P0 + P1 + P2 + … + P49

28,925 =

d = 21 = 273/13

P0 = 64 = 127 – 3(21)

LINEAR GROWTH WORD PROBLEMS: SOLUTIONS

14)  Starting in the year 2000, the number of crimes committed each year in is predicted to grow according to a linear growth model. During the year 2005, Monroe recorded 115 crimes. During the year 2006, Monroe recorded 121 crimes.

a.  How many crimes were committed in 2000?

85 crimes = 115 – 6(5)

b.  How many crimes are predicted to occur in 2007?

127 crimes = 121 + 6

c.  How many total crimes were committed between 2000 and 2010?

1265 =

d.  Write an equation for the number of crimes in years since 2000.

PN = 85 + 6N

15)  The production of toy trains is scheduled to be increased by 27 trains a month. It costs the company $3.50 to produce each train per month. When production started in cost the company $2800 for all trains.

a.  How many trains were produced initially?

800 trains = 2800/3.5

b.  How many trains will be produced in the 5th month of production?

935 trains = 800 + 27(5)

c.  Write an equation for the number of trains produced each month.

PN = 800 + 27N

d.  After one year, how many total trains were produced?

12,506 trains = (800 + 1124)*13/2

Initial Population = 800 and 12 months later = 1124

16)  A small business sells $15,500 worth of products during its first year of business. The owner has set an annual goal for increased sales at $1250 for 30 years. Find the total sales during just the first 12 years this business is in operation.

(P0 + P11)*12/2 = $268,500

SECTION 10.3 – EXPONENTIAL GROWTH MODELS- SOLUTIONS

For each of the following exponential growth model,

SOLUTIONS

1)  P0 = 12 and r = 3.5

PN = 12(3.5)N;

P9 = 945,787.66

2)  3, 6, 12, …

PN = 3(2)N

P9 = 1536

r = 6/3

3)  3.5, 9.45, 25.515, 68.8905, …

PN = 3.5(2.7)N;

P9 = 26689.59

r = 9.45/3.5 = 2.7

4)  P4 = 202.5 and P5 = 303.75

PN = 40(1.5)N;
P9 = 1537.734375

r = 303.75/ 202.5

P0 = 202.5/(1.5)4

For each of the following exponential growth model, Find given sum

5)  r = 2.2 and P0 = 3
P0 + P1 + P2 + … + P10

= 14,605.46

0 à 10 = 11 terms

6)  900 + 900(0.75) + 900(0.75)2 + … + 900(0.75)6

= 3119.458008

0 à 6 = 7 terms

7)  125 + 150 + 180 + … + 373.248

373.248 = 125(1.2)6 = P6

= 1614.49

0 à 6 = 7 terms

8)  r = 0.5 and P0 = 98304

8a. P0 + P1 + P2 + … + P13

= 196,596

0 à 13 = 14 terms

8b. P3 + P4 + … + P11

P3 = 98304(0.5)3 = 12288
= 24,528

3 à 11 = 9 terms

10.3 EXPONENTIAL GROWTH WORD PROBLEMS: SOLUTIONS

9)  The number of applicants is expected to grow by 35% each year for the next 15 years. If the original number of applicants was 220, then how many applicants are there predicted to be in 10 years?

100% + 35% = 135% à 1.35 = r

220(1.35)10 = 4424

10)  The number of reported cases of a virus is supposed to decay by 15% each year for 10 years. There currently are 1,200,000 cases of the virus. How cases are expected 6 years from now?

100% - 15% = 85% à 0.85 = r

1200000(0.85)6 = 452,579.4

11)  The number of certain type of bacteria increases at a rate of 20% every year. Suppose there were 3600 bacteria in 2009.

a.  How many bacteria were there in 2007?

3600/(1.2)2 = 2500 (divide by r to move backwards)

b.  How many bacteria will there be in 2012?

3600*(1.2)3 = 6221(multiply by r to move forwards)

c.  Write an equation for the number of bacteria per year since 2007.

PN = 2500(1.2)N; r = 1.2 and P0 = 2500 (year 2007)

12)  How much INTEREST would you earn on an account with 20% annual interest rate compounded daily that you initially invested $2500 after 3 years? k = 365(daily) i = 0.20 ß 20%

Interest = 4554.55 – 2500 = 2054.55

13)  Michaela has an option between two savings accounts. Account #1 she plans to invest $5500 at 20% annual interest rate compounded quarterly and Account #2 she plans to invest $6000 at 15% annual interest rate compounded monthly. Which account should she choose if she only needed to save money for 5 years?

Account #1 (k = 4, quarterly): 14,593.14

Interest = 9093.714

Account #2 (k = 12, monthly): 12,643.09

Interest = 6643.09

14)  What is the annual yield for an account with 10% annual interest compounded quarterly?

à 10.38%

SECTION 10.4: LOGISTIC GROWTH MODEL SOLUTIONS

a. Write TRANSITION RULE and SEED for the logistic growth model

b. Find p1, p2, and p3 for the given situation.

c. Determine the PREDICTED behavior

Stable Equilibrium, Cycle Pattern, Attracting Point or Chaos

If possible, provide the VALUE(S) of the predicted behavior

1) The initial p-value is p0 = 0.3 and r = 3.2.

a. pn = 3.2(1 – pn-1)( pn-1) Seed = 0.3

b. p1 = 0.972 p2 = 0.70533 p3 = 0.66509

c. 2- CYCLE: 0.513 and 0.799

2) The population of students is currently 1200. The maximum capacity of the school is actually 2000. The growth parameter of students is 1.5

a. pn = 1.5(1 – pn-1)( pn-1) Seed = 0.6 = 1200/2000

b. p1 = 0.36 p2 = 0.3456 p3 = 0.33924

c. Attracting Point: 0.333 or 1/3

3) p0 = 0.7 and r = 3.5.

a. pn = 3.5(1 – pn-1)( pn-1) Seed = 0.7

b. p1 = 0.735 p2 = 0.68171 p3 = 0.75943

c. 4- CYCLE: 0.382, 0.827, 0.500, 0.875

4) The population of salmon in a fish farm is 20,000 with a carrying capacity of 25,000 fish in the farm. The growth parameter of this species of salmon is 2.5.

a. pn = 2.5(1 – pn-1)( pn-1) Seed = 0.8 = 20000/25000

b. p1 = 0.4 p2 = 0.6 p3 = 0.6

c. 2- CYCLE: 0.513 and 0.799