Asked by Joe
                A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 59 cells. (The tolerance is 1% unless your instructor states otherwise.)
(a) Find the relative growth rate.
(b) Find an expression for the number of cells after t hours.
(c) Find the number of cells after 5 hours.
(d) Find the rate of growth after 5 hours.
            
        (a) Find the relative growth rate.
(b) Find an expression for the number of cells after t hours.
(c) Find the number of cells after 5 hours.
(d) Find the rate of growth after 5 hours.
Answers
                    Answered by
            Reiny
            
    Number = 59(2)^(t/20) where t is the number of minutes
so after 5 hours
Number = 59(2)^15
= 1933312
rate of growth
= d(Number)/dt
= (59ln2(2)^(t/20))/20
= (59/20)ln(2)^(t/20)
    
so after 5 hours
Number = 59(2)^15
= 1933312
rate of growth
= d(Number)/dt
= (59ln2(2)^(t/20))/20
= (59/20)ln(2)^(t/20)
                    Answered by
            unknown
            
    omg this is when i was bearly born
    
                    Answered by
            unknown
            
    this guy was like 12 now hes about 25
    
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