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Pre-Calc Homework Solutions 260

# Pre-Calc Homework Solutions 260 - 260 Section 6.4 11(a...

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9. (a) Annually: 2 5 1.0475 t ln 2 5 t ln 1.0475 t 5 } ln 1 ln .0 2 475 } < 14.94 years (b) Monthly: 2 5 1 1 1 } 0.0 1 4 2 75 } 2 12 t ln 2 5 12 t ln 1 1 1 } 0. 1 0 2 475 } 2 t 5 < 14.62 years (c) Quarterly: 2 5 1 1 1 } 0.0 4 475 } 2 4 t ln 2 5 4 t ln 1.011875 t 5 } 4 ln 1 ln .0 2 11875 } < 14.68 years (d) Continuously: 2 5 e 0.0475 t ln 2 5 0.0475 t t 5 } 0. l 0 n 4 2 75 } < 14.59 years 10. (a) Annually: 2 5 1.0825 t ln 2 5 t ln 1.0825 t 5 } ln 1 ln .0 2 825 } < 8.74 years (b) Monthly: 2 5 1 1 1 } 0.0 1 8 2 25 } 2 12 t ln 2 5 12 t ln 1 1 1 } 0.0 1 8 2 25 } 2 t 5 < 8.43 years (c) Quarterly: 2 5 1 1 1 } 0.0 4 825 } 2 4 t ln 2 5 4 t ln 1.020625 t 5 } 4 ln 1 ln .0 2 20625 } < 8.49 years (d) Continuously: 2 5 e 0.0825 t ln 2 5 0.0825 t t 5 } 0. l 0 n 8 2 25 } < 8.40 years 11. (a) Since there are 48 half-hour doubling times in 24 hours, there will be 2 48 < 2.8 3 10 14 bacteria. (b) The bacteria reproduce fast enough that even if many are destroyed there are still enough left to make the person sick. 12. Using y 5 y 0 e kt , we have 10,000 5 y 0 e 3 k and 40,000 5 y 0 e 5 k . Hence } 4 1 0 0 , , 0 0 0 0 0 0 } 5 } y y 0 0 e e 5 3 k k } , which gives e 2 k 5 4, or k 5 ln 2. Solving 10,000 5 y 0 e 3ln 2 , we have y 0 5 1250. There were 1250 bacteria initially. We could solve this more quickly by noticing that the population increased by a factor of 4, i.e. doubled twice, in 2 hrs, so the doubling time is 1 hr. Thus in 3 hrs the population would have doubled 3 times, so the initial
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