In 2006, the population of tewksbury, rhode island was 25,000, and it was growing at an annual rate of 2.2%.
what is the growth factor for the town?
write an equation to model tewksbury’s growth.
use your equation to estimate the population of tewksbury in 2011. round your response to the nearest whole number.

Assuming the growth is of 1st order, we can start using the formula for rate of 1st order reaction:

dN / dt = k * N

Rearranging,

dN / N = k dt

Where N = amount of sample, k = growth factor, t = time

Integrating the equation from N = Ni to Nf and t = ti to tf will result in:

ln (Nf / Ni) = k (tf – ti)

Finding for the growth factor k:

k = ln (Nf / Ni) / (tf – ti)

k = ln (1.022 Ni / Ni) / 1 year

k = 0.02176 / year

The population in 2011 is:

ln (Nf / Ni) = k (tf – ti)

ln (Nf / 25000) = (0.02176 / year) * (2011 – 2006)

Nf = 27,873.7 = 27,874

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

168

step-by-step explanation:

i am pretty sure that the answer is 168