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 61 cells. Find the relative growth rate.

the relative growth rate k = ln( 8 )

Step-by-step explanation:

Given the data in the question;

Initial population = 61 cells

which divides into two cells every 20 minutes;

Now.

dy/dt = ky

where y(t) = y(0)

y(t) is the population of bacteria

t is time in hours

y(0) is initial population

we know that, Initial population y(0) = 61

time t = 20 min = ( 20 / 60 )hr = 1/3 hr

so, y(1/3) = 2( 61 ) = 122

Now,

y(t) = y(0)

y(1/3) = 61

122 = 61

divide both sides by 122

2 =

ln(2) = k × 1/3

k = 3ln(2)

k = ln( 2³ )

k = ln( 8 )

Therefore, the relative growth rate k = ln( 8 )

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5