Suprinamaharjan95idn
Answered

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the present population of a certain place is 67205 the population growth rate is 5% per annum find the population of that place a year before​

Sagot :

Daichi1112008idn

Answer:

Let P be the population two years ago.

∴ P(1+

100

5

)

2

=104832

P(

100

105

)

2

=104832

P×

100

105

×

100

105

=104832

P=

105×105

104832×100×100

=95085.71

=95,086 (rounding off to the nearest whole number)

∴ Two years ago the population was 95,086.

Step-by-step explanation:

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https://youtu.be/PkKnp4SdE-w

Answer:

Step-by-step explanation:

∴ P(1+

100

5

)

2

=104832

P(

100

105

)

2

=104832

P×

100

105

×

100

105

=104832

P=

105×105

104832×100×100

=95085.71

=95,086 (rounding off to the nearest whole number)

∴ Two years ago the population was 95,086.

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