Answer:

So then P =11000 is the minimum that the least populated district could have.
Step-by-step explanation:
We have a big total of N = 132000 for the population.
And we know that we divide this population into 11 districts
And we have this info given "no district is to have a population that is more than 10 percent greater than the population of any other district"
Let's assume that P represent our minimum value for a district in the population. The range of possible values for the population of each district would be between P and 1.1 P
The interest on this case is find the minimum value for P and in order to do this we can assume that 1 district present the minimum and the other 10 the maximum value 1.1P in order to find which value of P satisfy this condition, and we have this:


So then P =11000 is the minimum that the least populated district could have.
Solution
Number of weeks in a year = 52 weeks
If in a normal year when each package of flea treatment lasts for 4 weeks, then in a year there Jim's dog will have to be treated for

Where as, when the fleas are bad in a year, the treatment lasts for only 3 weeks.
Then in a year Jim's dog would get

So Jim's dog will get 17- 13 =4 treatments more.
4 treatments that are made in 3 weeks each will be 4×3 =12 weeks more treatment
Answer:
Subtract 11 from both sides
Step-by-step explanation:
We need to find the expression for " number_of_prizes is divisible number_of_participants". Also there should not remain any remainder left. On in order words, we can say the reaminder we get after division is 0.
Let us assume number of Prizes are = p and
Number of participants = n.
If we divide number of Prizes by number of participants and there will be not remainder then there would be some quotient remaining and that quotent would be a whole number.
Let us assume that quotent is taken by q.
So, we can setup an expression now.
Let us rephrase the statement .
" Number of Prizes ÷ Number of participants = quotient".
p ÷ n = q.
In fraction form we can write
p/n =q ; n ≠ 0.