The population of a town increased by 16 2 3 % in 1990 and 10% in 1991. If the population of the town at the beginning of 1990 was 30,000, what was it at the end of 1991?

 38,500

Step-by-step explanation:

162/3% = (16 2/3)/100 = (50/3)/100 = 50/300 = 1/6

1/6 of the original population was ...

  1/6 × 30,000 = 5,000

So the population at the end of 1990 was

  30,000 +5,000 = 35,000.

In 1991, 10% of that number was added, so the population at the end of 1991 was ...

  35,000 + 0.10×35,000 = 38,500

38500 people

This is in my RSM homework.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2