At the end of the Year 1 the population will be = 110,000 + 4% of 110,000 =P1At the end of the Year 2 the population will be=P1+ 4% of P1= 110,000 + 4% of 110,000 + 4% of (110,000 + 4% of 110,000) = 110,000 + 4% of (110,000+ 110,000 + 4% of 110,000) = 110,000 + 2*4% of 110,000 + (4%)2 of 110,000= P2At the end of the Year 3 the population will be= P2+ 4% of P2= 110,000 +2* 4% of 110,000 + (4%)2 of 110,000 + 4% of [ 110,000 +2* 4% of 110,000 + (4%)2 of 110,000] = 110,000 +2* 4% of 110,000 + (4%)2 of 110,000 + 4% of 110,000 + 2* (4%)2 of 110,000 + (4%)3 of 110,000 =110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000 =P3At the end of the Year 4 the population will be= P3+ 4% of P3=110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000 +4% of [110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000] =110,000 +4* 4% of 110,000 + 6*(4%)2 of 110,000+4* (4%)3 of 110,000+ (4%)4 of 110,000 Now if we substitute n for 110,000 and r for 4% yearly rate of increase, we can rewrite,Before Year 1, at Year 0, the population, P0= n = n(1+r)0At the end of the Year 1 the population , P1= n+rn= x(1+r)1At the end of the Year 2 the population , P2= n+2rn+r2n =n(1+2r+r2)=n(1+r)2At the end of the Year 3 the population , P3= n+3rn+2r2n+r3n=n(1+3r+3r2+r3)=x(1+r)3At the end of the Year 4 the population , P4= n+4rn+6r2n+4r3n+r4n=n(1+r)4.
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At the end of the Year n the population , Px=n+nrx+(x-1)r2n+(x-2)r3n+...........+(x-x+2)r(x-1)x+(x-x+1)rxn=n(1+r)x.....At the end of the Year 16 the population , P16= n+16rn+15r2n+14r3n+.............+3r14n+2r15n+r16n=n(1+r)16 Thus under given condition of rate of growth, the Population P(x) at xth year will be P(x)=n(1+r)x Therefore, a population of 110,000 growing at 4% per year, in 16 years will be= 110,000(1+4/100)16= 110,000(1.04)16=206027.93702999115428132473599427≈206028
I hope this is helpful. I found the solution on the internet. Hope you have a good day and bye.
Answer:
M=-3d+36
Thank you! I hope you have fun in math, although personally I hate it.
Answer:
Total cost for tiles and paints is $924.
Step-by-step explanation:
We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.
The paint will be required for 4 walls and ceiling.
Let us find area of walls and ceiling.
Therefore, the area of walls and ceiling is 1410 square meters.
Given: Cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. Therefore,
Therefore, the total painting cost is $564.
Tiles will be required for floor. Let us find the area of floor.
Given: 1m squared floor tiles costs $3. So,
Therefore, the total tiles cost is $360.
Now let us find combined total cost of tiles and paint.
Therefore, the combined total cost of tiles and paint is $924.
