A typical middle-income household in 1980 earned $34,757. A similar household in 2009 earned $38,550. What was the relative increase in income for these households from 1980 to 2009? Round to the nearest one percent.

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chisnau chisnau · Answer 1 · 2019-02-20T06:51:06+01:00

Given conditions are :

In 1980's, a typical middle-income household earned= $34,757

In 2009, a similar middle-income household earned= $38,550

And we have to find relative increase in income for these households from 1980 to 2009.

So first we will find the total increase in amounts.

[tex]38550-34757=3793[/tex]

Relative increase = [tex]\frac{3793}{34757}*100[/tex]

= 10.91% or rounding it off we get approx 11%.

Hence, the answer is 11%.

lidaralbany lidaralbany · Answer 2 · 2019-07-20T03:41:10+02:00

The relative increase in income for these households from 1980 to 2009 is:

11%

If the initial amount is : a

and the increased amount the other year is: b

Then the relative increase is calculated by:

[tex]\dfrac{b-a}{a}[/tex]

and in percent we calculate it as:

[tex]\dfrac{b-a}{a}\times 100[/tex]

Here ,

A typical middle-income household in 1980 earned $34,757.

i.e. a=$ 34,757

A similar household in 2009 earned $38,550.

i.e. b=$ 38,550

then the relative increase in percent from 1980 to 2009 is:

[tex]=\dfrac{38,550-34,757}{34757}\times 100\\\\\\=\dfrac{3793}{34757}\times 100\\\\\\=0.1091291\times 100\\\\=10.91291\%[/tex]

Now, on rounding to the nearest percent we have:

Relative increase= 11%