It is estimated that t days after a virus begins to spread in a town, the percent of the population infected by the virus is approximated by p left parenthesis t right parenthesis equals t squared plus 5 tp(t)=t2+5t for 0 less than or equals t less than or equals 60≤t≤6. a. find the average rate of change of p with respect to t over the interval 1 to 55 days. b. find and interpret the instantaneous rate of change of p with respect to t at tequals=22.

The expression for p(t) can be easier to evaluate if it is factored to
   p(t) = t·(t+5)

a. The average rate of change (r) on the interval [0, 5] is
   r = (p(5) - p(0))/(5 - 0)
   r = (5·(5+5) - 0·(0+5))/5
   r = 10 . . . . percent per day

b. Differentiating the equation, you have
   p'(t) = 2t + 5
Evaluating that at t=2 gives
   p'(2) = 2·2 + 5 = 9 . . . . . percent per day, instantaneous rate at t=2

Interpretation: on day 2, the rate of infection is 9% of the population per day.

Its Easy All you need to do Make the Fraction into a Decimal and then you Compare like this Heres the Answer 
4 Divided by 5 = 0.8 Now the Question was >>>>>>    replace 4/5 to 0.8 ? 0.8 = 0.8 = 0.8
Hope it Helps Give Me a Thanks and Comment Me :D