Respuesta :
General formula to be used:Â [tex]F=P(1+i)^n[/tex]
F = future worth; P = present worth, i = annual interest, n = years
1.)Â [tex]F=5000(1+0.05)^{3.5} [/tex]
  F = $5931.06
2.) [tex]3000+262.50=3000(1+i)^{3.5} [/tex]
   i = 2.43%
3.) [tex]1300+299=1300(1+0.0575)^{n} [/tex]
   n = 3.7 = 4 years
4.) [tex]F=315(1+0.015)^{3} [/tex]
   F = $329.39
5.) [tex]P+5=P(1+0.0125)^{1212} [/tex]
   P = $1.45
   *I doubt the 1212 years. I think that is a typo error. Nevertheless, just      substitute the correct number of years to determine P :)
F = future worth; P = present worth, i = annual interest, n = years
1.)Â [tex]F=5000(1+0.05)^{3.5} [/tex]
  F = $5931.06
2.) [tex]3000+262.50=3000(1+i)^{3.5} [/tex]
   i = 2.43%
3.) [tex]1300+299=1300(1+0.0575)^{n} [/tex]
   n = 3.7 = 4 years
4.) [tex]F=315(1+0.015)^{3} [/tex]
   F = $329.39
5.) [tex]P+5=P(1+0.0125)^{1212} [/tex]
   P = $1.45
   *I doubt the 1212 years. I think that is a typo error. Nevertheless, just      substitute the correct number of years to determine P :)