Honors Discrete Chapter 10 Test Review Guide

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					              Honors Discrete Chapter 10 Test Practice Problems
SECTION 10.2
For each of the following linear growth model,
    Find an explicit description for the population
    Find the 9th generation of the population, P9

1) P0 = 718 and d = 32                                   4) P13 = 217 and P14 = 228
PN = 718 + 32N; P9 = 1006                                PN = 74 + 11N; P9 = 173
2) 3, 19, 35, 51, …                                      5) P6 = 165 and P15 = 318
PN = 3 + 16N; P9 = 147                                   PN = 63 + 17N; P9 = 216
3) P5 = 97 and P6 = 86                                   6) P21 = 1896 and P90 =4449
PN = 152 - 11N; P9 = 53                                  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                 10) 17 + 29 + 41 + … + 1157
232 = (8 + 50)/2 * 8                                    56,352 = (17 + 1157)/2 * 96
8) P0 = 115 and d = 83;                                 11) 8895 + 8778 + … + 7140
    a. P0 + P1 + P2 + … + P29                           128,280 = (8895 + 7140)/2 * 16
    39,555 = (115 + 2522)/2 * 30                        12) P3 = 127 and P16 = 400;
    b. P21 + … + P31                                        a. P0 + P1 + P2 + … + P49
    25,003 = (1858 + 2688)/2 * 11                           28,925 = (64 + 1093)/2 * 50
    c. P125 + … + P157                                      b. P0 + P1 + … + P19 + P39 + P40 + … + P62
    389,994 = ( 10,490 + 13,146)/2 * 33                     32,258 = (64+ 463)/2 * 20 + (883+ 1366)/2 * 14
9) For 50 terms, 37 + 44 + 51+ …                            c. P2 + P4 + P6 + … + P22 + P24 + P26
     10,425 = (37+ 380)/2 * 50                              4654 = (106 + 610)/2 * 13
WORD PROBLEMS:
13) 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
    b. How many crimes are predicted to occur in 2007? 127 crimes
    c. How many total crimes were committed between 2000 and 2010? 1265
    d. Write an equation that describes the number of crimes in years since 2000. PN = 85 + 6N


14) 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
    b. How many trains will be produced in the 5th month of production? 935 trains
    c. Write an equation that describes the number of trains produced each month. PN = 800 + 27N
    d. Write an equation that describes the production cost of trains produced each month.
        PN = 2800 + 94.5N
    e. After one year, how much money did the company devote to producing trains? 43,771
    f. After one year, how many total trains were produced? 12,506 trains

15) 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. Assuming the goal is met, find the total sales
    during the first 12 years this business is in operation. (P0 + P11)*12/2 = $268,500
SECTION 10.3
For each of the following exponential growth model,
     Find an explicit description for the population
     Find the 9th generation of the population, P9
1) P0 = 12 and r = 3.5                                    4) P5 = 99 and P6 = 90
             N
PN = 12(3.5) ; P9 = 945,787.66                            PN = 159.44(10/11)N; P9 = 67.62
2) 3, 6, 12, …                                            5) P7 = 1200 and P9 = 1875
          N
PN = 3(3) ; P9 = 1536                                     PN = 251.66(1.25)N; P9 = 1875
3) 3.5, 9.45, 25.515, 68.8905, …                          6) P3 = 2500 and P6 = 8437.5
PN = 3.5(2.7)N; P9 = 26689.59                             PN = 740.74(1.5)N; P9 = 28476.56
For each of the following exponential growth model,
     Find the given sum
7) r = 2.2 and P0 = 3                           9) r = .5 and P0 = 98304
    P0 + P1 + P2 + … + P10                         a. P0 + P1 + P2 + … + P13 98304(.514-1)/(.5-1) = 196,596
3(2.211-1)/(2.2-1) = 14,605.46                     b. P7 + P8 + … + P15 768(.59-1)/(.5-1) = 1533
8) 125 + 150 + 180 + … + 373.248                   c. P3 + P4 + … + P1112,288(.59-1)/(.5-1) = 24,528
        7
125(1.2 -1)/(1.2-1) = 1614.49                      d. P1 + P3 + P5 + P7 + P9 + P11 + P13
                                                   49,152(.257-1)/(.25-1) = 65,532
WORD PROBLEMS:
10) 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?
    220(1.35)10 = 4424
11) The number of reported cases of a virus is suppose 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?
    1,200,000(.85)10 = 452,579.4
12) 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? 2500
        b. How many bacteria there be in 2012? 6221
        c. Write an equation that describes the number of bacteria per year since 2007. PN = 2500(1.2)N

13) How much interest would you earn on an account with 20% annual interest rate compounded daily
                                                                  3* 365
                                                               .2 
    that you initially invested $2500 after 3 years? 2500 1        2500  2054.55
                                                              365 
14) 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.
    a. Which account should she choose if she only needed to save money for 5 years?
        Account #1: 14,593 v. Account #2: 12,643
    b. Which account should she choose if she needed to save money for 10 years?
        Account #1: 38,720 v. Account #2: 22,952

15) Determine the annual yield for a savings account with 15% annual interest rate compounded
                           4
                    .15 
    quarterly.  1        1 =15.87%
                     4 
16) Determine the annual yield if you originally invest $300 and 2 years later the account has $376.32.
300r2 = 376.32, r = 1.12, 12%

SECTION 10.4: Please see class work problems, quiz review guide, and quiz from Section 10.4 for similar
problems to what you will see on the test. Make sure you have become proficient at using your calculator
to help predict long-term behavior of a logistic growth model.

				
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