Auranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. if the population at the end of 1999 was 9,320:

Step-by-step explanation:

Heres the complete question:

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. If the population at the end of 1999 was 9,320:

How many people lived in the town at the beginning of 1985? (Round your answer to the nearest whole number.)

solution:

Let the population of the town at the beginning of 1985 be P. Then, given that in the first five-year period the population declined by 3.2%, i.e., 0.032, the population of the town at the end of 1989 would be

(1 – 0.032)P = 0.968P.

Again, given that in the second five-year period the population declined by 5.2%, i.e., 0.052, the population of the town at the end of 1994 would be

(1 – 0.052)(0.968P) = 0.948 x 0.968P = 0.917664P.

Finally, given that in the third five-year period the population declined by 4.7%, i.e., 0.047, the population of the town at the end of 1999 would be

(1 – 0.047)(0.917664P) = 0.874533792P.

We are given, 0.874533792P = 9320 or

P = 9320/0.874533792 = 10657.11.

Thus, 10657 people lived in the town at the beginning of 1985

answer: a. 74

step-by-step explanation:

it shows divided evenly, so add the 37+37 = 74

you never say the total so we cant find the percentage used for petrol

step-by-step explanation: