{"id":58,"date":"2025-01-02T23:05:04","date_gmt":"2025-01-02T23:05:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/introduction-to-modeling-get-stronger-answer-key\/"},"modified":"2025-01-02T23:05:04","modified_gmt":"2025-01-02T23:05:04","slug":"introduction-to-modeling-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/introduction-to-modeling-get-stronger-answer-key\/","title":{"raw":"Introduction to Modeling: Get Stronger Answer Key","rendered":"Introduction to Modeling: Get Stronger Answer Key"},"content":{"raw":"\n

1.<\/p>\n

    \n\t
  1. [latex]P_0=20.[\/latex] [latex]P_n=P_{n\u22121}+5[\/latex]<\/li>\n\t
  2. [latex]P_n=20+5n[\/latex]<\/li>\n<\/ol>\n

    3.<\/p>\n

      \n\t
    1. [latex]P_1=P_0+15=40+15=55.[\/latex] [latex]P_2=55+15=70[\/latex]<\/li>\n\t
    2. [latex]P_n=40+15n[\/latex]<\/li>\n\t
    3. [latex]P_{10}=40+15(10)=190[\/latex] thousand dollars<\/li>\n\t
    4. [latex]40+15n=100[\/latex] when [latex]n=4[\/latex] years.<\/li>\n<\/ol>\n

      5. Grew [latex]64[\/latex] in [latex]8[\/latex] weeks: [latex]8[\/latex] per week<\/p>\n

        \n\t
      1. [latex]P_n=3+8n[\/latex]<\/li>\n\t
      2. [latex]187=3+8n.[\/latex] [latex]n=23[\/latex] weeks<\/li>\n<\/ol>\n

        7.<\/p>\n

          \n\t
        1. [latex]P_0=200[\/latex](thousand), [latex]P_n=(1+.09)[latex]P_{n\u22121}[\/latex] where [latex]n[\/latex] is years after 2000<\/li>\n\t
        2. [latex]P_n=200(1.09)^n[\/latex]<\/li>\n\t
        3. [latex]P_{16}=200(1.09)^{16}=794.061[\/latex](thousand)[latex]=794,061[\/latex]<\/li>\n\t
        4. [latex]200(1.09)^n=400. n=log(2)\/log(1.09)=8.043[\/latex]. In 2008<\/li>\n<\/ol>\n

          9. Let [latex]n=0[\/latex] be 1983. [latex]P_n=1700(2.9)^n[\/latex]. 2005 is [latex]n=22[\/latex]. [latex]P_{22}=1700(2.9)^{22}=25,304,914,552,324[\/latex] people. Clearly not realistic, but mathematically accurate.<\/p>\n

          11. If [latex]n[\/latex] is in hours, better to start with the explicit form. [latex]P_0=300.[\/latex] [latex]P_4=500=300(1+r)^4[\/latex]
          \n[latex]500\/300=(1+r)^4.[\/latex]
          \n[latex]1+r=1.136.[\/latex]
          \n[latex]r=0.136[\/latex]<\/p>\n

            \n\t
          1. [latex]P_0=300. P_n=(1.136)P_{n\u22121}[\/latex]<\/li>\n\t
          2. [latex]P_n=300(1.136)^n[\/latex]<\/li>\n\t
          3. [latex]P_{24}=300(1.136)^24=6400[\/latex] bacteria<\/li>\n\t
          4. [latex]300(1.136)n=900.n=log(3)\/log(1.136)=[\/latex] about [latex]8.62[\/latex] hours<\/li>\n<\/ol>\n

            13.<\/p>\n

              \n\t
            1. [latex]P_0=100[\/latex] [latex]P_n=P_{n\u22121}+0.70(1\u2212P_{n\u22121}\/2000)P_{n\u22121}[\/latex]<\/li>\n\t
            2. [latex]P_1=100+0.70(1\u2212100\/2000)(100)=166.5[\/latex]<\/li>\n\t
            3. [latex]P_2=166.5+0.70(1\u2212166.5\/2000)(166.5)=273.3[\/latex]<\/li>\n<\/ol>\n

              15. To find the growth rate, suppose [latex]n=0[\/latex] was [latex]1968[\/latex]. Then [latex]P_0[\/latex] would be [latex]1.60[\/latex] and [latex]P2=2.30=1.60(1+r)^8,r=0.0464[\/latex]. Since we want [latex]n=0[\/latex] to correspond to 1960, then we don't know [latex]P_0[\/latex], but [latex]P_8[\/latex] would be [latex]1.60=P_0(1.0464)^8[\/latex]. [latex]P_0=1.113[\/latex]<\/p>\n

                \n\t
              1. [latex]P_n=1.113(1.0464)^n[\/latex]<\/li>\n\t
              2. [latex]P_0=$1.113[\/latex], or about [latex]$1.11[\/latex]<\/li>\n\t
              3. 1996 would be [latex]n=36.[\/latex] [latex]P_36=1.113(1.0464)^36=$5.697[\/latex]. Actual is slightly lower.<\/li>\n<\/ol>\n

                17. The population in the town was [latex]4000[\/latex] in 2005, and is growing by [latex]4\\%[\/latex] per year.<\/p>\n","rendered":"

                1.<\/p>\n

                  \n
                1. [latex]P_0=20.[\/latex] [latex]P_n=P_{n\u22121}+5[\/latex]<\/li>\n
                2. [latex]P_n=20+5n[\/latex]<\/li>\n<\/ol>\n

                  3.<\/p>\n

                    \n
                  1. [latex]P_1=P_0+15=40+15=55.[\/latex] [latex]P_2=55+15=70[\/latex]<\/li>\n
                  2. [latex]P_n=40+15n[\/latex]<\/li>\n
                  3. [latex]P_{10}=40+15(10)=190[\/latex] thousand dollars<\/li>\n
                  4. [latex]40+15n=100[\/latex] when [latex]n=4[\/latex] years.<\/li>\n<\/ol>\n

                    5. Grew [latex]64[\/latex] in [latex]8[\/latex] weeks: [latex]8[\/latex] per week<\/p>\n

                      \n
                    1. [latex]P_n=3+8n[\/latex]<\/li>\n
                    2. [latex]187=3+8n.[\/latex] [latex]n=23[\/latex] weeks<\/li>\n<\/ol>\n

                      7.<\/p>\n

                        \n
                      1. [latex]P_0=200[\/latex](thousand), [latex]P_n=(1+.09)[latex]P_{n\u22121}[\/latex] where [latex]n[\/latex] is years after 2000<\/li>\n
                      2. [latex]P_n=200(1.09)^n[\/latex]<\/li>\n
                      3. [latex]P_{16}=200(1.09)^{16}=794.061[\/latex](thousand)[latex]=794,061[\/latex]<\/li>\n
                      4. [latex]200(1.09)^n=400. n=log(2)\/log(1.09)=8.043[\/latex]. In 2008<\/li>\n<\/ol>\n

                        9. Let [latex]n=0[\/latex] be 1983. [latex]P_n=1700(2.9)^n[\/latex]. 2005 is [latex]n=22[\/latex]. [latex]P_{22}=1700(2.9)^{22}=25,304,914,552,324[\/latex] people. Clearly not realistic, but mathematically accurate.<\/p>\n

                        11. If [latex]n[\/latex] is in hours, better to start with the explicit form. [latex]P_0=300.[\/latex] [latex]P_4=500=300(1+r)^4[\/latex]
                        \n[latex]500\/300=(1+r)^4.[\/latex]
                        \n[latex]1+r=1.136.[\/latex]
                        \n[latex]r=0.136[\/latex]<\/p>\n

                          \n
                        1. [latex]P_0=300. P_n=(1.136)P_{n\u22121}[\/latex]<\/li>\n
                        2. [latex]P_n=300(1.136)^n[\/latex]<\/li>\n
                        3. [latex]P_{24}=300(1.136)^24=6400[\/latex] bacteria<\/li>\n
                        4. [latex]300(1.136)n=900.n=log(3)\/log(1.136)=[\/latex] about [latex]8.62[\/latex] hours<\/li>\n<\/ol>\n

                          13.<\/p>\n

                            \n
                          1. [latex]P_0=100[\/latex] [latex]P_n=P_{n\u22121}+0.70(1\u2212P_{n\u22121}\/2000)P_{n\u22121}[\/latex]<\/li>\n
                          2. [latex]P_1=100+0.70(1\u2212100\/2000)(100)=166.5[\/latex]<\/li>\n
                          3. [latex]P_2=166.5+0.70(1\u2212166.5\/2000)(166.5)=273.3[\/latex]<\/li>\n<\/ol>\n

                            15. To find the growth rate, suppose [latex]n=0[\/latex] was [latex]1968[\/latex]. Then [latex]P_0[\/latex] would be [latex]1.60[\/latex] and [latex]P2=2.30=1.60(1+r)^8,r=0.0464[\/latex]. Since we want [latex]n=0[\/latex] to correspond to 1960, then we don't know [latex]P_0[\/latex], but [latex]P_8[\/latex] would be [latex]1.60=P_0(1.0464)^8[\/latex]. [latex]P_0=1.113[\/latex]<\/p>\n

                              \n
                            1. [latex]P_n=1.113(1.0464)^n[\/latex]<\/li>\n
                            2. [latex]P_0=$1.113[\/latex], or about [latex]$1.11[\/latex]<\/li>\n
                            3. 1996 would be [latex]n=36.[\/latex] [latex]P_36=1.113(1.0464)^36=$5.697[\/latex]. Actual is slightly lower.<\/li>\n<\/ol>\n

                              17. The population in the town was [latex]4000[\/latex] in 2005, and is growing by [latex]4\\%[\/latex] per year.<\/p>\n","protected":false},"author":6,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":3,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/58"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/58\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/58\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=58"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=58"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=58"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=58"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}