Question

suppose the population of a town is 100,000 in 1999. The population increased by 4.5% each year. what will be the population in 2005?

Answer 1

u need to take out 5 % of 100000 for the first year...
than take out 5% OF THE PREVIOUS YEAR ADDED WITH THE RISE IN POPULATION...
DO THIS 5-6 TIMES UPTIL 2005

Answer 2

still confused

Answer 3

this is the formula A = P(1+r)^t
to
A = 100000(1+0.045)^6

Answer 4

do u get it?

Answer 5

255972?

Answer 6

no

Answer 7

ur number is going to be lower

Answer 8

remember PEMDAS

Answer 9

well thats what i got using that formula

Answer 10

do u want to know the answer?
first do parentheses did u do that?

Answer 11

here hold on

Answer 12

yes

Answer 13

A = P(1+r)t (t is as an exponent)
In the formula A = P(1+r)t, P is the principal\, r is the annual rate of interest, and A is the amount after t years. An account earning interest at a rate of 4% has a principal of $500,000. If no more deposits or withdrawls are made, about how much money will be in the account after five years?
A $705,200
B $620,700
C $608,300
D $575,000
try this the answer is
A = 500000(1.04)^5
A = 500000*1.2166529
A = $608,326.45, so guess it's c

Answer 14

I based the formula on the compound interest formula which is Total amount = (Beginning amount) * (1 + rate)^years

Answer 15

if u give up tell me..

Answer 16

hello?

Answer 17

here:
Population = 100,000 *1.045^n (n is number of years since 1999. So, the population in 2005 (where n=6) is 130,226

Answer 18

so the answer is right above this message